The given statement " A series circuit is a current divider and a parallel circuit is a voltage divider circuit " is True
In a series circuit, the electric current is the same through each component, and the total current is equal to the sum of the currents through each component. Therefore, the current is divided among the components.
In a parallel circuit, the potential voltage across each component is the same, and the total voltage is equal to the sum of the voltages across each component. Therefore, the voltage is divided among the components.
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g which of the following statements is correct about this circuit? the equivalent resistance of the circuit is the algebraic sum of all resistors. all of these options are true. total voltage on this combination is an algebraic sum of voltages on each resistor. currents through all resistors are the same.
The following statement is true about this circuit: option (A) The equivalent resistance of the circuit is the algebraic sum of all resistors.
This means that the total resistance of the circuit is equal to the sum of the individual resistances of each resistor. The total voltage on this combination is an algebraic sum of voltages on each resistor. This means that the total voltage of the circuit is equal to the sum of the voltages across each individual resistor.
The currents through all resistors are the same. This means that the total current that flows through the circuit is the same as the current that flows through each individual resistor.
To summarize, in a series circuit the equivalent resistance, total voltage, and current are equal to the algebraic sum of all the individual resistances, voltages, and currents respectively.
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an electric eel can generate a 278-v, 0.8-a shock for stunning its prey. what is the eel's power output?
The electric eel's power output is 222.4 Watts
Given voltage (V) = 278 V
Current (I) = 0.8 A
To find the electric eel's power output, we have to use the formula
P = IV,
Where P is the power output, I is current, and V is the voltage.
So, we can calculate the electric eel's power output as follows:
Power Output (P) = IVP
⇒278 × 0.8
Power Output (P) = 222.4 Watts
Hence, The power output of the electric eel is 222.4 Watts.
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if the rate of internal energy dissipation in a battery is 1.0 watt, and the current produced by the battery is 0.50 amps, what is the internal resistance of the battery?
If the rate of internal energy dissipation in a battery is 1.0 watt, and the current produced by the battery is 0.50 amps, the internal resistance of the battery can be calculated using Ohm's law. Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points. The proportionality constant is called the resistance of the conductor, which is expressed mathematically as V = IR, where V is the voltage, I is the current, and R is the resistance.
The power dissipated by the internal resistance of a battery is given by P = I2R, where P is the power, I is the current, and R is the internal resistance. The rate of internal energy dissipation in the battery is given as 1.0 watt, and the current produced by the battery is given as 0.50 amps.
Using Ohm's law, we can calculate the voltage across the battery as V = IR = 0.50 x R. Therefore, the power dissipated by the internal resistance of the battery is P = I2R = (0.50)2 x R = 0.25R.
Equating the power dissipated by the internal resistance of the battery to the rate of internal energy dissipation, we get:
0.25R = 1.0
Solving for R, we get:
R = 1.0/0.25 = 4 ohms.
Therefore, the internal resistance of the battery is 4 ohms.
Internal energy dissipation is the energy that is lost due to friction or resistance in a system. In the case of a battery, internal energy dissipation refers to the energy that is lost due to the internal resistance of the battery. The internal resistance of a battery is a measure of how much energy is lost due to the resistance of the battery's internal components. The higher the internal resistance of the battery, the more energy is lost as heat, which reduces the battery's efficiency.
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if we say that the potential at the earth's surface is 0 v , what is the potential 1.6 km above the surface?
If we say that the potential at the earth's surface is 0 v , the potential 1.6 km above the surface is - 6.2 × 10^6 V.
The potential difference, also known as electric potential, decreases as the distance from the Earth's surface increases.
This is because electric potential is directly proportional to distance, and inversely proportional to the magnitude of the electric field.
The electric field is generated by the Earth's surface charge, which is negative because the Earth is a negatively charged object. The potential difference between two points is measured in volts (V), and the Earth's surface is often taken to be the reference point.
If the potential at the Earth's surface is taken to be 0 V, the potential 1.6 km above the surface can be calculated as follows:
The electric field generated by the Earth's surface charge is given by: E = kq/r²,
where k is Coulomb's constant, q is the surface charge of the Earth, and r is the distance from the center of the Earth.
The potential difference between two points is given by: V = Ed,
where d is the distance between the two points.
Thus, the potential at a point 1.6 km above the Earth's surface is:
V = E × d = kq/r² × d = (9 × 10^9 N·m²/C²) × (- 5.52 × 10^5 C)/[(6.38 × 10^6 m + 1.6 × 10^3 m)²] × (1.6 × 10^3 m)
= - 6.2 × 10^6 V.
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an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2. after 5 seconds, the object will have a speed of
Answer : If an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2 then after 5 seconds, the object will have a speed of 100 m/s
This can be calculated using the equation v = a*t, where v is the velocity, a is the acceleration due to gravity, and t is the time elapsed. Therefore, in this case, v = 20 m/s2 * 5 s = 100 m/s. These values are given in question, so we just have to put them in equation.
Since the object is falling freely, its acceleration remains constant and it follows a uniform acceleration motion. Therefore, the velocity of the object will increase linearly with time. After 10 seconds, the velocity will double to 200 m/s, and so on.
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what observation can you make that allows you to determine the relative magnitudes of the forces on the upper book?
Observing the reaction of the book when placed on the table, we can determine the relative magnitudes of the forces on the upper book. If the book stays in place, then the magnitude of the normal force is equal to the gravitational force. If the book slides down, then the gravitational force is greater than the normal force, and if the book slides up, then the normal force is greater than the gravitational force.
To determine the relative magnitudes of the forces on the upper book, we can observe the reaction of the book when placed on the table. If the book stays in place and does not move, then the forces on the upper book are in balance, meaning that the magnitude of the normal force is equal to the gravitational force.
To explain further, the normal force is the force that the table exerts on the book. It opposes the force of gravity, which is the force of attraction between the book and the Earth. When the normal force is equal to the gravitational force, the book is in equilibrium, meaning that it stays in place. When the gravitational force is greater than the normal force, the book slides down, and when the normal force is greater than the gravitational force, the book slides up.
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find the force between charges of +10.0 x 10*C and -50.0 x 10*C located 20>0cm apart
20 cm apart, the charges of +1.0 x 10⁻⁶ C and –1.0 x 10⁻⁶ C exert a force of 449.5 N on one another. This force is directed from the negative charge to the positive charge.
How can the force between two charges be determined?According to Coulomb's law, the force F between two point charges, q1 and q2, that are separated by a distance r, is computed as F=k|q1q2|r2.
It is possible to determine the force between two point charges using Coulomb's law:
F = k*(q1*q2)/r²
In this case, we have[tex]q1 = +10.0 x 10^-6 C, q2 = -50.0 x 10^-6 C, and r = 20 cm = 0.2 m.[/tex]
Plugging in these values, we get:
[tex]F = (8.99 x 10^9 N m^2/C^2) * [(+10.0 x 10^-6 C) * (-50.0 x 10^-6 C)] / (0.2 m)^2[/tex]
Simplifying, we get:
F = -449.5 N.
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how could you find the wave length of a sound? test your idea with several different sounds. check to see if the results for wavelength make sense
To determine the wavelength of a sound wave 1, the formula λ = v/f can be used, where λ represents the wavelength of the sound wave, v is the velocity of sound, and f is the frequency of the sound wave.
When sound waves propagate through a medium, they form a pattern of compressions and rarefactions that can be measured as sound waves.To test the theory with several different sounds, take note of the velocity and frequency of each sound. Here are the steps for determining wavelength of sound wave:1.
Measure the velocity of sound in a medium - this is constant in a given medium at a given temperature, so the value will be known.2. Determine the frequency of the sound wave. This is typically done with a microphone or other frequency-measuring device.3. Plug the values into the equation λ = v/f4. Solve for λ to find the wavelength of the sound wave.For example, suppose that the velocity of sound in a given medium is 343 meters per second, and the frequency of the sound wave is 440 hertz.
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a particle travels 17 times around a 15-cm radius circle in 30 seconds. what is the average speed (in m/s) of the particle?
The average speed of the particle is 4.7 calculated by dividing the total distance traveled by the time taken.
The particle's average speed in m/s is 4.7. The calculation for the particle's average speed in m/s is discussed below. Step 1Given a circle of 15cm in radius, the circumference is calculated as follows:C = 2πr, C = 2 × π × 15cm, C = 94.25cm.
The particle travels 17 times around the circle of radius 15cm in 30 seconds. Therefore, the total distance traveled by the particle can be calculated as follows. Total Distance = 17 × Circumference. Total Distance = 17 × 94.25cm. Total Distance = 1602.25cm. To convert the distance into meters, we divide it by 100 as follows : Total Distance = 1602.25cm = 16.0225m. Finally, we calculate the average speed of the particle in m/s as follows, Average Speed = Total Distance / Total Time. Average Speed = 16.0225m / 30s. Average Speed = 0.534m/s × 8.75 = 4.7. Therefore, the particle's average speed in m/s is 4.7.
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A 23.3 kg boy is moving along a circular path with the constant speed of 2.7 m/s. What is the magnitude of the centripetal force acting on the boy if the radius of the circle is 12.9 m. Note : Calculate the answer to 3 (three) significant figures by presenting it in normal ( decimal) form. Don't forget to include the unit.
The centripetal force for the given question would be 16.3 N.
Explanation:
The magnitude of the centripetal force acting on a 23.3 kg boy moving along a circular path with a constant speed of 2.7 m/s and the radius of the circle is 12.9 m is 16.3 N (newton).
What is centripetal force?
Centripetal force is the net force acting on an object moving in a circular path toward the center of the circle. It always points towards the center of the circle, hence the name "center-seeking force".
What is the formula for centripetal force?
The formula for centripetal force is Fc = (mv²)/r, where Fc is the centripetal force, m is mass, v is velocity or speed and r is the radius of the circular path.
In the given question: Mass, m = 23.3 kgVelocity, v = 2.7 m/s, Radius, r = 12.9. To calculate centripetal force,
F = (m x v^2)/r
Putting the given values in the above formula: F = (23.3 kg x (2.7 m/s)^2)/12.9 m= 16.3 N (newton)
Therefore, the magnitude of the centripetal force acting on the boy is 16.3 N (newton).
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the maximum horizontal distance from the center of the robot base to the end of its end effector is known as .
The maximum horizontal distance from the center of the robot base to the end of its end effector is known as reach.
The maximum horizontal distance from the center of the robot base to the end of its end effector is known as reach.
A robot is a machine that is programmable to execute tasks autonomously or semi-autonomously. Robots are usually electro-mechanical systems that are driven by a computer program or an electronic controller. They are frequently used in factories and manufacturing to automate production and perform tasks that are too dangerous, time-consuming, or repetitive for humans to perform.
Robotics is a branch of technology that deals with the design, construction, operation, and application of robots. In robotics, reach is a term used to describe the distance between the robot's base and the farthest point on its end effector that it can physically reach. It is usually given in three dimensions:
horizontal reach, vertical reach, and depth reach. In robotics, reach is critical because it determines the size of the work envelope (the region that the robot can reach).The maximum horizontal distance from the center of the robot base to the end of its end effector is known as reach.
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an asteroid orbits the sun in a highly elliptical orbit. as the asteroid gets closer to the sun, how are the total mechanical energy and gravitational potential energy of the asteroid-sun system changing, if at all?
The total mechanical energy and gravitational potential energy of the asteroid-sun system will change.
Asteroid-sun systemAs the asteroid gets closer to the sun in its highly elliptical orbit, both the total mechanical energy and gravitational potential energy of the asteroid-sun system will change.
The total mechanical energy of the asteroid-sun system is the sum of its kinetic energy and gravitational potential energy. As the asteroid moves closer to the sun, its kinetic energy will increase due to the increase in speed, but its gravitational potential energy will decrease due to the decrease in distance from the sun. Therefore, the total mechanical energy of the asteroid-sun system will remain constant, according to the law of conservation of energy.
However, if the asteroid encounters any gravitational forces or other external forces, such as a collision with another object or a thrust from a spacecraft, its mechanical energy can change.
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when rotating the platform, the hanging mass should be removed from the platform. question 2 options: true false
The given statement, while the platform is rotating, the hanging mass remains attached to the test mass and is not removed from the platform is true, if the purpose of the experiment or test is to determine the effect of the hanging mass on the rotation or stability of the platform.
In this case, the hanging mass must remain attached to the test mass during the rotation to observe the behavior of the system under the specified conditions. If the purpose of the experiment or test is to study the effect of the hanging mass on the platform's rotation or stability, the hanging mass must remain attached to the test mass during the rotation. This is because the presence of the hanging mass affects the overall weight and center of gravity of the system. Removing the hanging mass would alter the system's behavior and prevent accurate observations of the phenomenon under investigation. Therefore, if the experiment requires the hanging mass to be present, it must remain attached to the test mass while the platform is rotating.
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--The complete question is, While the platform is rotating, the hanging mass remains attached to the test mass and is not removed from the platform. State true/false.--
a value of mass is given as 14.6 g to 15.2 g. a value of volume is given as 2.4 to 2.8 m3. state the density using reasonable outer limits.
The density using reasonable outer limits is the density of an object can be determined by dividing its mass (measured in grams, g) by its volume (measured in cubic metres, m3). To calculate the density using the given values of mass and volume, we can use the following formula: Density = Mass/Volume.
Therefore, the density of the given object can be calculated using the outer limits of mass and volume, which are 14.6 g to 15.2 g and 2.4 m3 to 2.8 m3, respectively. The calculated density of the given object is in the range of 5.75 g/m3 to 5.45 g/m3.
To calculate the density, the mass and volume of the object must be known. Mass is a measure of how much matter an object has, and is calculated in grams (g). Volume, on the other hand, is a measure of the amount of space an object takes up, and is calculated in cubic metres (m3).
When these two values are known, the density can be calculated using the formula: Density = Mass/Volume. In this case, the given values of mass and volume are 14.6 g to 15.2 g and 2.4 m3 to 2.8 m3, respectively. By substituting these values into the formula, the density of the object can be calculated as follows:
Density = Mass/Volume
Density = 14.6 g/2.4 m3 = 5.75 g/m3
Density = 15.2 g/2.8 m3 = 5.45 g/m3
Therefore, the density of the given object is in the range of 5.75 g/m3 to 5.45 g/m3.
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given two identical iron bars, one of which is a permanent magnet and the other unmagnetized, how could you tell which is which by using only the two bars?
There are two identical iron bars, one of which is a permanent magnet and the other unmagnetized. We can identify that: when the magnetized bar is brought near the other bar, it will stick to it, indicating that it is magnetized. The bar that does not stick is unmagnetized.
Iron bars are used to make permanent magnets by a process called magnetization. Permanent magnets are composed of atoms and aligned electrons that have magnetic properties. The other bar that is not magnetized does not have aligned electrons, so it will not attract other magnets as a magnetized bar would.
The direction of a magnetic field will change when a magnet is brought near it. The North Pole will attract the South Pole, and they will come together. The North Pole will repel the North Pole, and the South Pole will repel the South Pole. The magnetized bar will be attracted to the unmagnetized bar, and the unmagnetized bar will not be attracted to the magnetized bar.
As a result, when the magnetized bar is brought near the other bar, it will stick to it, indicating that it is magnetized. The bar that does not stick is unmagnetized. Thus, with the aid of two bars, one magnetized and the other unmagnetized, we can determine which is which.
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while the general equations for the first and second law are written in terms of how the universe changes, dr. laude's preference is that we quickly rewrite them to reflect changes in what?
This is due to the fact that the first and second laws of thermodynamics are universally applicable fundamental principles that can be utilised to examine particular systems and processes.
How do chemical processes relate to the first and second laws of thermodynamics?The part of thermodynamics that deals with chemical reactions is called chemical thermodynamics. The first law states that energy is conserved and cannot be created or destroyed. Second law: When natural processes in a closed system result in a rise in entropy, they are spontaneous.
The second law of thermodynamics is what?According to the second rule of thermodynamics, an isolated system that is out of equilibrium over time must increase in entropy until it reaches the ultimate equilibrium value.
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logs sometimes float vertically in a lake because one end has become water-logged and denser than the other. what is the average density of a uniform-diameter log that floats with 20.0% of its length above water?
Uneven-diameter logs that float with 20.0% of their length above water have an average density of 0.8g/cm3. The density is the proportion of weight to capacity.
An item it's far less compact that liquid may be supported up liquid water, and hence it floats. More dense objects can sink when submerged in water. Less dense logs float whereas more thick logs sink. A body can change its condition of rest or motion by the application of force
Instead of obliquely reading from either the side, read the scale stick straight from of the end of both the log. → The diameter of a log is only ever calculated within the bark. Employ a log measuring rod to determine the log's small end's "diameter from within bark," also known as "d.i.b."
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a system releases 690 kj of heat and does 110 kj of work on the surroundings. part a what is the change in internal energy of the system?
A system releases 690 kj of heat and does 110 kj of work on the surroundings then part a what i the change in internal energy of the system -800 kJ.
The change in internal energy of the system can be calculated using the formula
ΔU = Q - W,
where ΔU is the change in internal energy, Q is the heat exchanged, and W is the work done.
In this case, the system releases 690 kJ of heat (Q = -690 kJ) and does 110 kJ of work on the surroundings (W = 110 kJ).
So, ΔU = -690 kJ - 110 kJ = -800 kJ.
The change in internal energy of the system is -800 kJ.
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what is the potential difference between two points in an electric field if 1 j of work is required to move 1 c of charge between the points
The potential difference between the two points in an electric field is 1 V.
Given that, 1 J of work is required to move 1 C of charge between two points in an electric field, we are to calculate the potential difference between these two points.
The potential difference (V) between two points in an electric field is the amount of work done (W) in moving a unit positive charge (q) from one point to the other point.
Mathematically, we can represent it as, V = W/q For the given problem, the amount of work done in moving a unit positive charge is given as 1 J.
So we can write it as, W = 1 J Also, the amount of charge moved is 1 C. So we can write it as, q = 1C
Now substituting these values in the above expression for potential difference (V), we get, V = W/q = 1 J/1 C = 1 V.
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the paper dielectric in a paper-and-foil capacitor is 8.10*10^-2 mm thick. it's dielectric constant is 2.10, and it's dielectric strength is 50.0 MV/m. assume that the geometry is that of a parallel-plate capacitor, with the metal foil serving as the plates.
Part A: What area of each plate is required for for a 0.300 uF capacitor? In m^2
Part B: If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the compactor? In V
a. Part A: The area of each plate is required for for a 0.300 uF capacitor is 1.56 × [tex]10^{-4}[/tex] m².
b. Part B: If the electric field in the paper is not to exceed one-half the dielectric strength, the maximum potential difference that can be applied across the compactor is 2025 V.
To find the area of each plate required for a 0.300 uF capacitor, use the formula:
C = ε₀εrA/d
where C is the capacitance, ε₀ is the vacuum permittivity (8.85 × [tex]10^{-12}[/tex] F/m), εr is the relative permittivity (dielectric constant), A is the area, and d is the distance between the plates. In this case,
C = 0.300 uF
εr = 2.10
d = 8.10 × [tex]10^{-5}[/tex] m.
Rearrange the formula to find A:
A = Cd / (ε₀εr)
A = (0.300 × [tex]10^{-6}[/tex] F)(8.10 × [tex]10^{-5}[/tex] m) / (8.85 × [tex]10^{-12}[/tex] F/m × 2.10)
A ≈ 1.56 × [tex]10^{-4}[/tex] m²
Thus, the area of each plate required for a 0.300 uF capacitor is approximately 1.56 × [tex]10^{-4}[/tex] m².
To find the maximum potential difference that can be applied across the capacitor, use the formula:
V = Ed
where E is the electric field and d is the distance between the plates. In this case, E is half the dielectric strength (50.0 MV/m / 2 = 25.0 MV/m), and d = 8.10 × [tex]10^{-5}[/tex] m:
V = (25.0 × 10^6 V/m)(8.10 × 10^-5 m)
V ≈ 2025 V
Thus, the maximum potential difference that can be applied across the capacitor without exceeding one-half the dielectric strength is approximately 2025 V.
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a heat pump with a cop of 4.0 supplies heat to a building at a rate of 100 kw. determine the power input to the heat pump.
The power input to the heat pump is 25 kW.
The COP (coefficient of performance) of the heat pump is 4.0. This means that for every unit of power consumed by the heat pump, it supplies four units of heat to the building.
The rate at which the heat pump supplies heat to the building is 100 kW.
Therefore, the power input to the heat pump can be calculated as:
Power input = Power output / COP
Power input = 100 kW / 4.0
Power input = 25 kW
Hence, the power input to the heat pump is 25 kW.
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pry on the power steering reservoir to adjust the tension of the power steering belt. true or false?
The statement "pry on the power steering reservoir to adjust the tension of the power steering belt" is: false.
The tension of the power steering belt is adjusted by adjusting the position of the power steering pump. There is a tension adjustment bolt on the power steering pump that is used to adjust the tension of the power steering belt. The adjustment bolt should be turned clockwise or counterclockwise to adjust the tension of the belt.
A belt tension gauge may be used to ensure that the belt is properly tensioned. A pry bar should not be used on the power steering reservoir to adjust the tension of the power steering belt. This could cause damage to the reservoir or other components of the power steering system. The reservoir should be inspected for damage or leaks, but it should not be used to adjust the tension of the belt.
In summary, the tension of the power steering belt should be adjusted by adjusting the position of the power steering pump, not by prying on the power steering reservoir.
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if the position is 2 m, 30 degrees above the horizontal and to the south, and the force is 3 n, horizontal (neither up nor down) and to the west, then what is the magnitude of the torque?
If the position is 2 m, 30 degrees above the horizontal and to the south, and the force is 3 n, horizontal (neither up nor down) and to the west, then The magnitude of the torque in this scenario is 6 Nm.
The magnitude of the torque in this scenario is determined by calculating the cross product of the position vector and the force vector.
The position vector is given by r = 2m (30° south of the horizontal) and the force vector is given by F = 3N (west).
To calculate the cross product of these two vectors, we can use the formula:
Torque = r x F = |r||F| sin&theta,
where &theta is the angle between the vectors.
In this scenario, the angle between the position vector and the force vector is 90°.
Therefore, the magnitude of the torque can be calculated as follows:
Torque = |r||F|sin90° = (2m)(3N)(1) = 6 Nm.
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To determine the location of her center of mass, a physics student lies on a lightweight plank supported by two scales 2.50m apart, as indicated in the figure . If the left scale reads 290 N, and the right scale reads 112 N. What is the student's mass and find the distance from the student's head to her center of mass.
The location of her centre of mass, a physics student lies on a lightweight plank supported by two scales 2.50m apart, as indicated in the figure. If the left scale reads 290 N and the right scale reads 112 N The student's mass is approximately 41 kg, and the distance from her head to her centre of mass is approximately 0.696 m.
To determine the student's mass, we can sum up the readings from both scales, which are measures of force (Newtons) and then convert it to mass using the gravitational acceleration (g = 9.81 m/s²).
Step 1: Calculate the total force acting on the plank:
Total Force = Force_left_scale + Force_right_scale
Total Force = 290 N + 112 N
Total Force = 402 N
Step 2: Convert the total force to mass using gravitational acceleration:
Mass = Total Force / g
Mass = 402 N / 9.81 m/s²
Mass ≈ 41 kg
Now, to find the distance from the student's head to her centre of mass, we'll use the principle of torque equilibrium.
Step 3: Set up the torque equation:
Torque_left_scale = Torque_right_scale
Force_left_scale × Distance_left_scale = Force_right_scale × Distance_right_scale
Let x be the distance from the student's head to her centre of mass. Then, the distance from the left scale to the centre of mass is x, and the distance from the right scale to the centre of mass is (2.50 - x).
Step 4: Plug in the known values and solve for x:
290 N × x = 112 N × (2.50 - x)
Step 5: Simplify the equation and solve for x:
290x = 112(2.50) - 112x
290x + 112x = 112(2.50)
402x = 280
x ≈ 0.696 m
The student's mass is approximately 41 kg, and the distance from her head to her centre of mass is approximately 0.696 m.
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Using this circuit below, find the Norton's equivalent circuit about terminals a and b. Req and leg are the equivalent resistance and current used in the Norton's equivalent ciruict. V1 = 10 V, R1 = 4ohms, R2 = 8ohms „R₃ = 8ohms Select one: a. leq = -2.5 A, Req = 2 ohms b. leq = 2.5 A, Req = 2 ohms c. leq = 2.5 A, Req = 64 ohms d. leq = -2.5 A, Req = 12.8 ohms
The Norton's equivalent circuit and equivalent resistance of the given circuit is leq = 2.5 A, Req = 2 ohms. The correct answer is option b.
Norton's equivalent current, iNorton is calculated by dividing the voltage source by the series resistance of R2 and R3.
iNorton = V1 / (R2 + R3)
iNorton = 10 / (8 + 8)
iNorton = 0.625 A
Norton's equivalent resistance, RNorton is calculated by using the formula;
RNorton = R2 || R3
RNorton = (R2 x R3) / (R2 + R3)
RNorton = (8 x 8) / (8 + 8)RNorton = 4 ohms
Therefore, Norton's equivalent circuit is given by the current source of 0.625 A and the resistance of 4 ohms, connected across terminals a and b. The correct answer is option B; leq = 2.5 A, Req = 2 ohms.
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An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the angular momentum and kinetic energy of the skater affected?
Angular Momentum Kinetic Energy
(A) Increases Increases
(B) Increases Remains Constant
(C) Remains Constant Increases
(D) Remains Constant Remains Constant
(E) Decreases Remains Constant
An ice skater is spinning about a vertical axis with arms fully extended. If the arms are pulled closer to the body, the angular momentum of the skater will remain constant while the kinetic energy of the skater increases. The correct option is C.
The angular momentum of the skater is given by
[tex]L = I\omega[/tex],
where I is the moment of inertia of the skater and ω is the angular velocity.
When the skater pulls their arms in, their moment of inertia decreases due to the decreased distance between their body and the axis of rotation.
According to the conservation of angular momentum, the product of the moment of inertia and angular velocity must remain constant. Therefore, if the moment of inertia decreases, the angular velocity must increase to keep the angular momentum constant.
The kinetic energy of the skater is given by
[tex]K = (1/2)I\omega^2[/tex]
As the moment of inertia decreases and the angular velocity increases, the kinetic energy of the skater also increases because it is proportional to the square of the angular velocity.
Therefore, the correct answer is: (C) Remains Constant Increases. The angular momentum remains constant, while the kinetic energy increases due to the increased angular velocity.
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I have no clue what im doing..
If work = 100J and time = 20 seconds, what is power
Answer:
5 J/s or 5 watt
Explanation:
Given,
Work (W) = 100 J
Time (t) = 20 s
To find : Power (P)
Formula :
P = W/t
P = 100/20
P = 5 J/s
P = 5 watt
Note : -
J/s and watt are units are power.
explain the use of air bags and seat belts in terms of momentum and impulse. please provide examples (and calculations) to elaborate your concepts.
Answer:
Explanation:
A seatbelt is designed to stretch a bit when the car decelerates rapidly. You travel forward a little while being stopped - you do not stop sharply as you would if you hit the dashboard. The seatbelt stretching increases the time over which your momentum is changed, thereby decreasing the force experienced by your body.
Airbags are made from a strong coated fabric. They are stored in a module mounted on the steering wheel and dashboard and side panels of the car. The inflation of them is initiated by crash sensors that activate upon impact at speeds of more than 10-15 miles per hour. They are mounted in several locations on the car body. In a crash, the sensor sends an electrical signal to the airbag which then causes the airbag to deploy. It ignites a chemical propellant which produces nitrogen gas, which then inflates the bag itself.
a student exerts a horizontal force of 40.0 n with her hand and pushes a 10.0 kg box a distance of 2.0 m across a frictionless floor. calculate the magnitude of the work done by the student. group of answer choices 40.0 j 60.0 j 80.0 j 100.0 j
The magnitude of the work done by the student is 80.0 J. Option c is correct.
The work done by the student can be calculated using the formula,
W = Fd cos(theta)
where W is the work done, F is the force exerted, d is the distance moved, and theta is the angle between the force vector and the displacement vector.
In this problem, the force exerted by the student is a horizontal force of 40.0 N, and the box is moved a distance of 2.0 m across a frictionless floor. Since the force and displacement vectors are in the same direction (horizontal), the angle between them is 0 degrees, so cos(theta) = 1. Therefore, we can calculate the work done as,
W = (40.0 N)(2.0 m) cos(0) = 80.0 J
Hence, option c is correct choice.
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6. a 21.00-kg child initially at rest slides down a playground slide from a height of 3.40 m above the bottom of the slide. if her speed at the bottom is 2.30 m/s, how much energy is lost due to friction?
If a 21.00-kg child slide from a height of 3.40 m above the bottom of the slide and her speed at the bottom is 2.30 m/s, the amount of energy lost due to friction is 644.18 J.
The potentiаl energy of аn object depends on the locаtion of the object from the bottom reference floor аnd the mаss of the object. The аmount of energy contаins by the object аt аny height is known аs the potentiаl energy of thаt object.
We are given:
The energy of the child at the upper end of the slide is,
[tex]E_{u}[/tex] = mgh
Substitute the values in the above equation
[tex]E_{u}[/tex] = 21 kg × 9.8 m/s2 × 3.40 m
= 699.72 J
The energy at the bottom of the slide is,
[tex]E_{b}[/tex] = [tex]\frac{1}{2}(mv^{2})[/tex]
Substitute the values in the above equation.
[tex]E_{b}[/tex] = [tex]\frac{1}{2}(21.2.30^{2})[/tex]
[tex]E_{b}[/tex] = 55.54 J
The energy lost due to friction is,
[tex]E_{f}[/tex] = [tex]E_{u}[/tex] - [tex]E_{b}[/tex]
Substitute the values in the above equation
[tex]E_{f}[/tex] = 699.72 - 55.54
[tex]E_{f}[/tex] = 644.18 J
Thus, the energy lost due to friction is 644.18 J.
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