a metal object is suspended from a spring scale. the scale reads 920 n when the object is suspended in air, and 750 n when the object is completely submerged in water. a. draw a diagram showing the three forces acting on the submerged object. b. find the volume of the object. c. find the density of the metal.

William Herschel's approach failed due to the fact that some parts of the Milky Way galaxy are denser than others.

This means that the number of stars would be greater in these regions, making it difficult to determine the galaxy's center simply by counting the number of stars in different directions. Herschel's pioneering work, including his discovery of Uranus and his cataloging of hundreds of nebulae, helped pave the way for future astronomers to explore and understand the universe. However, his method for locating the center of the Milky Way was limited by the technology of his time.

In modern times, astronomers have employed a range of techniques to study the galaxy, including measuring the positions and motions of stars, observing the behavior of gas and dust clouds, and using radio and other wavelengths of light to observe the galaxy's structure and composition.

Despite these advances, the center of the Milky Way remains difficult to observe directly due to the presence of dense dust and gas clouds, which block visible light. Nonetheless, astronomers have been able to estimate the location and size of the galaxy's central region through careful analysis of the behavior of stars and other objects orbiting around its center.

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When a block is tied to a post with a cable and rotating with a constant velocity, on a horizontal smooth surface, the direction of its acceleration is towards the center of rotation.

Acceleration is a vector quantity that represents a change in velocity in terms of magnitude and direction. When an object changes direction, it is accelerating, and its direction of acceleration is perpendicular to its direction of motion. When an object rotates with a constant velocity, its speed remains constant, but its direction changes continuously. As a result, it is continuously accelerating towards the center of rotation, as in the case of a block tied to a post with a cable rotating on a horizontal smooth surface.

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The given statement, "piezoelectricity is a property where quartz crystals vibrate 100,000 times a second if heated to 100 degrees Celsius" is false because piezoelectricity is a property of certain materials, including quartz crystals, that generates an electric charge in response to mechanical stress or pressure, not heat.

Piezoelectricity is a property of certain materials, including quartz crystals, that generates an electrical voltage in response to mechanical stress or pressure. Heating quartz crystals to 100 degrees Celsius does not cause them to vibrate 100,000 times per second, although it may affect their piezoelectric properties in other ways. The frequency of vibration for a quartz crystal oscillator is determined by its physical dimensions and properties, and may be in the range of thousands or millions of vibrations per second, depending on the design and application of the oscillator.

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The angular momentum of a 2.9-kg uniform cylindrical grinding wheel of radius 28 cm when rotating at 1500 rpm is 1.18 kg m²/s. the angular momentum of the grinding wheel is 14.5 kg m²/s.

The formula for angular momentum is:

L = Iω

Where, L = angular momentum

I = moment of inertia

ω = angular velocity

First, we need to find the moment of inertia of the grinding wheel.

The moment of inertia of a uniform cylinder is given by:

I = (1/2)mr²

Where,m = mass of the cylinder (2.9 kg)

r = radius of the cylinder (28 cm = 0.28 m)

So, I = (1/2)(2.9 kg)(0.28 m)²

     I = 0.092 kg m²

Now, we can find the angular momentum:

L = Iω

ω = angular velocity = 1500 , rpm = 157.08 rad/s (1 revolution = 2π radians, so 1500 rpm = 1500/60 = 25

revolutions per second = 25 × 2π = 157.08 radians per second)

L = (0.092 kg m²)(157.08 rad/s)L

  = 14.5 kg m²/s.

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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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Because both waves travel at the same speed, we know they must have distinct frequencies in order to have different wavelengths. This is due to the fact that the speed of light in a particular medium is constant,

and the frequency of a wave is inversely related to its wavelength. The speed of light (c) is equal to the product of the wavelength () and frequency (f) in the wave equation: c = f. Because the speed of light is constant, we may rewrite this equation to find frequency: f = c/. We have the following for wave a with a wavelength of 235 nm:

[tex]f_a = \frac{3.00 * 10^8 m/s}{235 * (10-9) m} = 1.28 x 10^{15} Hz[/tex]

We have the following for wave b with a wavelength of 515 nm:

[tex]f_b[/tex] = c / λ b  = 3.00 x 10⁸ m/s / (515 x 10⁻⁹ m) = 5.83 x 10¹⁴ Hz

Therefore, we can see that wave a has a higher frequency than wave b.

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The 73 kg man would need to run at approximately 5.92 m/s to have the same kinetic energy as an 8.0 g bullet fired at 400 m/s.

The kinetic energy (KE) of an object is given by the formula KE = (1/2)mv^2, where m is the mass of the object and v is its velocity.

To calculate the velocity of the 73 kg man, we can set his kinetic energy equal to that of the 8.0 g bullet, which is:

[tex]KE_bullet = (1/2)mv^2 = (1/2)(0.008 kg)(400 m/s)^2 = 640 J[/tex]

Now we can solve for the velocity (v) of the 73 kg man by setting his kinetic energy equal to 640 J:

[tex]KE_man = (1/2)mv^2 = 640 J(1/2)(73 kg)v^2 = 640 Jv^2 = 640 J x 2 / 73 kgv^2 = 35.068v = sqrt(35.068) = 5.92 m/s[/tex]

Therefore, the 73 kg man would need to run at approximately 5.92 m/s (21.3 km/h or 13.2 mph) to have the same kinetic energy as an 8.0 g bullet fired at 400 m/s.

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Answer:  The charge on the capacitor increases exponentially as the capacitor charges. As time goes on, the rate of charging decreases, and the charge on the capacitor approaches Qmax. The charge on the capacitor does not change once it is fully charged.

An uncharged capacitor is connected in series with a battery and a resistor. When the circuit is closed, the current begins to flow, and the capacitor begins to charge. The voltage across the capacitor increases as the capacitor charges.

When a battery, resistor, and uncharged capacitor are connected in series, the charge of the capacitor changes as a function of time according to the equation:

Q = Qmax(1 - e^(-t/RC))

An uncharged capacitor is connected in series with a battery and a resistor. When the circuit is closed, the current begins to flow, and the capacitor begins to charge. The voltage across the capacitor increases as the capacitor charges.

When the voltage across the capacitor is equal to the battery voltage, the current stops flowing through the circuit. The capacitor is then fully charged, and the charge on the capacitor is Qmax. At this point, the voltage across the capacitor is equal to the battery voltage, and the current through the resistor is zero.

The charge on the capacitor, Q, changes as a function of time, t, according to the equation:

Q = Qmax(1 - e^(-t/RC))

where Qmax is the maximum charge on the capacitor, R is the resistance of the resistor, C is the capacitance of the capacitor, and e is the base of natural logarithms.

The charge on the capacitor increases exponentially as the capacitor charges. As time goes on, the rate of charging decreases, and the charge on the capacitor approaches Qmax. The charge on the capacitor does not change once it is fully charged.



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In the cylinder, This equation can be solved for ω, and then v can be found using the relationship v = r * ω.

When the cylinder is released from rest, its gravitational potential energy is converted into kinetic energy (translational and rotational). To find the instantaneous velocity (v) and angular velocity (ω) of the cylinder, we can apply the conservation of mechanical energy and the relationship between linear and angular velocities.
Initially, the cylinder has potential energy (PE) due to its height (h) above the ground:
PE_initial = m * g * h
When the cylinder descends and starts rotating, it has both translational kinetic energy (KE_trans) and rotational kinetic energy (KE_rot):
KE_trans = 0.5 * m * v^2
KE_rot = 0.5 * I * ω^2
Since the string does not slip, we can relate linear velocity (v) to angular velocity (ω) as:
v = r * ω
Now, applying the conservation of mechanical energy:
PE_initial = KE_trans + KE_rot
Substituting the expressions for PE_initial, KE_trans, and KE_rot, and the relationship between v and ω, we get:
m * g * h = 0.5 * m * (r * ω)^2 + 0.5 * I * ω^2

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The kinetic energy of the snowball just before it hits the ground is 19.6 joules.

To determine the kinetic energy of the snowball just before it hits the ground, we can use the formula for kinetic energy:

KE = 1/2 m v²

where KE is the kinetic energy, m is the mass of the snowball, and v is the velocity of the snowball just before it hits the ground.

In this case, we know that the mass of the snowball is 0.80 kg and the velocity just before it hits the ground is equal to the initial velocity with which it was thrown (7.0 m/s) since air resistance is assumed to be negligible. Therefore, we can substitute these values into the formula:

KE = 1/2 * 0.80 kg * (7.0 m/s)²

KE = 19.6 J

Therefore, the kinetic energy of the snowball just before it hits the ground is 19.6 joules.

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Correct question is:

You throw a 0.80kg snowball at 7.0m/s straight down off a 5.0m tall bridge then what is the kinetic energy of the snowball just before it hits the ground?

The ratio of m1 and m2 is 2.81.

The ratio of m1 and m2 can be calculated using Newton's second law of motion, which states that the force F applied to an object is equal to its mass multiplied by its acceleration: F = m * a.

For the two objects given in the question, we have F = m1 * 7.36 m/s2 and F = m2 * 2.62 m/s2.

Therefore, the ratio of m1 and m2 can be found by dividing the first equation by the second: m1/m2 = (m1 * 7.36 m/s2) / (m2 * 2.62 m/s2). Solving for m1/m2, we get m1/m2 = 2.81.

The ratio of m1 and m2 is equal to 2.81, which can be calculated using Newton's second law of motion. According to the equation, the force F applied to an object is equal to its mass multiplied by its acceleration.

For the two objects in the question, we found the ratio of m1 and m2 to be 2.81.

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The teledeltos paper does not conduct charges well, in comparison the metal electrodes are better conductors. When the power supply is turned on,  charge builds up on the conductors. The path do charges follow between the electrodes is least resistance. The charges are transported from one electrode to the other via the path of least resistance.

Electrons have a negative charge, and they are attracted to a positive charge. As a result, electrons flow from negative to positive in electrical circuits. The movement of electrons is referred to as an electrical current. The charges flow through the conductor as the voltage is applied, and the path of least resistance is followed between the electrodes.

The charges travel through the metal electrodes because they have low resistance, while the teledeltos paper has a high resistance which means that it does not conduct charges well. The charges are transported through the metal electrodes between the two electrodes due to their excellent electrical conductivity.

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If you open the faucet further and water enters the tub at a higher rate, the water level in the tub will: rise

The water level will increase at a faster pace, and the tub will fill up more quickly than before. This happens because the rate of water flow into the tub is now higher than the rate at which it can drain away. Therefore, opening the faucet further increases the flow of water into the tub, which raises the water level at a higher rate.

The faucet opening determines the water flow rate, and the flow rate affects the filling rate of the tub. Thus, a higher flow rate leads to a higher filling rate of the tub. As a result, the water level in the tub increases more quickly when the faucet is opened further. The pressure of the incoming water is a critical factor in determining the rate at which the water fills up the tub.

When you turn the faucet on all the way, it releases the highest possible amount of water pressure into the tub, causing the water level to rise rapidly. In summary, opening the faucet further and letting water enter the tub at a higher rate will increase the water level in the tub, and the tub will fill up more quickly than before.

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The time it takes for the ball to reach the ground is 1.63 seconds.
The magnitude of the final velocity of the ball is 17.2 m/s.

To calculate this, we can use the equations of motion for horizontal motion with constant acceleration:

x = x0 + v0t + (1/2)at2

v2 = v02 + 2a(x - x0)

Here, x

is the initial velocity (17.2 m/s), x is the final distance (11.0 m), and a is the acceleration due to gravity (-9.8 m/s).
Substituting in the given values, we get:

11.0 m = 2.0 m + (17.2 m/s)(t) + (-9.8 m/s2)(t2)/2

(17.2 m/s)2 = (17.2 m/s)2 + 2(-9.8 m/s2)(11.0 m - 2.0 m)
Since the initial velocity was directed horizontally, the magnitude of the final velocity is the same as the initial velocity (17.2 m/s).

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To calculate the centripetal acceleration in m/s2 at the tip of a 3.50-meter-long helicopter blade that rotates at 300 rev/min, the given values should be converted into suitable units.

Then, we can use the following formula:Centripetal acceleration = (angular velocity)2 (radius)The conversion factor for rpm (rev/min) to rad/s is 2π/60 radians/second.

Therefore,Angular velocity = (300 rev/min)(2π/60) = 31.42 rad/sRadius = 3.50 centripetal acceleration = (31.42 rad/s)2 (3.50 m)= 3476 m/s2Therefore, the centripetal acceleration at the tip of a 3.50-meter-long helicopter blade that rotates at 300 rev/min is 3476 m/s2.

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Uranium belongs to the f-block of the periodic table. The correct option is fourth.

The f-block is located at the bottom of the periodic table, and it consists of the lanthanide and actinide series. Uranium is an actinide element, which means it is part of the second row of the f-block. It is widely used in nuclear power plants, as well as in nuclear weapons.

The f-block elements are known for their unique electron configurations, which include partially filled f-orbitals. These elements are also called "inner transition metals" because they fill their d-orbitals before filling their f-orbitals. Uranium is a radioactive metal that has 92 protons in its nucleus.

In summary, uranium belongs to the f-block of the periodic table, specifically the actinide series.

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The idealised ammeter and the 3 ohm resistor are connected in series, and they both get the same current. As a result, 2.13 A is likewise the idealized ammeter's reading.

An perfect ammeter's internal resistance is zero, whereas an ideal voltmeter's internal resistance is infinite.

The following formula can be used to get the parallel resistors' equivalent resistance:

1/Req = 1/12 + 1/9

1/Req = 3/36 + 4/36

1/Req = 7/36

Req = 36/7 ≈ 5.14 ohms

Now that the circuit has the equivalent resistance, we can redisplay it:

The circuit's overall current is determined by:

I = V / (Rint + Req)

I = 18 / (3.26 + 5.14)

I ≈ 2.13 A.

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A), Multiply the length of the vibrating string (60.0 cm) by the ratio to find the distance x. B)No, it's impossible to play a G4 note (392 Hz) on this string without retuning, C) not possible without retuning.

Part A) To find the distance x from the bridge to play a D5 note (587 Hz), follow these steps:
1. Calculate the speed of the wave on the string using the formula: v = √(T/μ), where T is tension and μ is linear mass density.
2. Calculate the wavelength of the A4 note using the formula: λ = v/f, where f is the frequency of the A4 note (440 Hz).
3. Calculate the wavelength of the D5 note using the formula: λ = v/f, where f is the frequency of the D5 note (587 Hz).
4. Find the ratio between the A4 and D5 wavelengths: λ_A4 / λ_D5.
5. Multiply the length of the vibrating string (60.0 cm) by the ratio to find the distance x.
Part B) No, it's impossible to play a G4 note (392 Hz) on this string without retuning.
Part C) The reason why it's impossible to play a G4 note (392 Hz) without retuning is because the frequencies of the fundamental modes are fixed and cannot be changed unless the tension, mass, or length of the string is altered. To play a G4 note, the string would need to be adjusted so that its fundamental frequency is 392 Hz, which is not possible without retuning.

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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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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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To get a capacitance of 3.00 pF with a plate separation of 4.00 mm and air between the plates, the plate area required is 1.062 × 10⁻⁵ m² (to 3 significant figures).

The plate separation, d = 4 mm. The capacitance, C = 3 pF = 3 × 10⁻¹² F.

We need to find the plate area, If the material between the plates is air, then the capacitance of a parallel plate capacitor can be given as:

[tex]$$C = \frac{\varepsilon_0A}{d}$$[/tex]

where, ε0 = permittivity of free space = 8.854 × 10⁻¹² F/m.

Substituting the given values in the above formula, we get:

[tex]$$\begin{aligned}C &= \frac{\varepsilon_0A}{d}\\ 3 × 10^{-12} &= \frac{8.854 × 10^{-12} \text{ F/m} × A}{4 × 10^{-3} \text{ m}}\\ A &= \frac{3 × 4 × 10^{-3} \text{ m} × 8.854 × 10^{-12} \text{ F/m}}{8.854 × 10^{-12} \text{ F/m} × 10^{-12}}\\ &= 1.062 × 10^{-5} \text{ m}^2 \end{aligned} $$[/tex]

Therefore, the plate area required to provide a capacitance of 3.00 pF is 1.062 × 10⁻⁵ m² (to three significant figures).

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The given values in the equation, F = (0.0147) * (3.09)/0.110F = 0.414 N Thus, the force required to give an angular acceleration of 3.09 m/s² to the solid ball is 0.414 N.

Given, The radius of a solid ball (r) = 0.110 m The mass of the solid ball (m) = 1.88 kg The angular acceleration of the solid ball (α) = 3.09 m/s²Now, we need to find the force required to give an angular acceleration of 3.09 to the solid ball. So, we will use the formula for torque, Torque (τ) = Fr Where, r = radius of the solid ball F = force required to move the solid ball on the edge of the solid ball By using Newton's second law of motion, F = ma Where, m = mass of the solid ball a = angular acceleration of the solid ball By using the formula for torque, Torque (τ) = Frτ = IαWhere, I = moment of inertia of the solid ball By equating both equations, F * r = IαF = Iα/r By using the formula for moment of inertia of a solid ball, I = (2/5)mr²I = (2/5) * 1.88 * 0.110²I = 0.0147 kg m²Now, substituting the given values in the equation, F = (0.0147) * (3.09)/0.110F = 0.414 N Thus, the force required to give an angular acceleration of 3.09 m/s² to the solid ball is 0.414 N.

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Etoposide, sold under the trade name Etopophos, is used for the treatment of lung cancer, testicular cancer, and lymphomas. To select all the O atoms that are part of the acetals in etoposide,

1. Look for acetal functional groups in the etoposide molecule.

Acetals consist of a central carbon atom bonded to two alkoxy (OR) groups and two alkyl (R) groups.
2. Identify the oxygen atoms that are part of these acetal groups.

Upon examination of the etoposide structure, you will find that there are two acetal functional groups present. The oxygen atoms that are part of the acetals in etoposide are those directly bonded to the central carbon atom in each acetal group.

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When a skydiver descends towards the earth with her parachute open, the work done by the drag force from the air is negative.

When a skydiver descends towards the earth with her parachute open, the drag force works in the opposite direction of the skydiver's motion, slowing her descent. The skydiver's motion is downward, whereas the drag force is upward. As a result, the angle between the drag force and the skydiver's motion is 180 degrees.

Because of the dot product, the work done by the drag force is negative.Work, which is a scalar quantity, is given by the following equation:

Work done = Force * Displacement * cos(θ)

where: θ is the angle between the applied force and the displacement vector. The work done is negative in this case because the angle between the applied force and the displacement is 180 degrees.

As a result, cos(180) is -1. This negative value results in the work done by the drag force from the air being negative.

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The spring constant is 0.28 N/m if 14 oscillations take 11.0s.

In physics, oscillations are defined as a repetitive variation, typically in time, of some measure about a central value (often a point of equilibrium) or between two or more different states. The spring constant (k) is a measure of a spring's stiffness. It is the amount of force required to displace a spring a specific distance (typically one meter).

The spring constant formula is expressed as:-

F=kx

where k is the spring constant and x is the displacement produced by the force F.

We know that the air-track glider has a mass of 240 g, and it is attached to a spring. The glider is pushed 8.2 cm against the spring and then released. The oscillations are then observed, and it is found that 14 oscillations occur in 11.0 s.

We can calculate the spring constant by using this information.

Let us now calculate the spring constant k.

For a mass (m) attached to a spring, the formula for the time period T is:-

T=2π√m/k

We know that the time period T = 11/14 s. We also know that the glider has a mass of 240 g or 0.24 kg.

Now, we can solve the formula for the spring constant k as follows:-

k= 4π²m/T²k = 4π² × 0.24 kg / (11/14 s)²k = 0.28 N/m

Therefore, the spring constant is 0.28 N/m.

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A cable that weighs 4 lb/ft is used to lift 550 lb of coal up a mine shaft 550 ft deep. The work done is 302500 joules (J).

Given the following data:

A cable that weighs 4 lb/ft is used to lift 550 lb of coal up a mine shaft 550 ft deep.

The formula to calculate the work done is,

Work Done (W) = Force (F) × Distance (D)

Where, Force (F) = Weight of Coal lifted, Distance (D) = Height of mine shaft

We are supposed to find the work done.

Hence, we will substitute the values in the above formula to calculate the work done.

W = 550 × 550W

= 302500 Units of Work

The units of work is in lb-ft which is equivalent to joules.

Hence the work done is 302500 joules (J).

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The position vector of the particle at an arbitrary time is vt.

Step by step explanation:

The position vector of the particle at an arbitrary time is a vector that has both direction and magnitude.

It is defined by its starting point and its endpoint.

Given that a particle passes through the point at time t, moving with constant velocity v, the position vector of the particle at an arbitrary time is given by the formula;

Position vector of the particle = Position vector of the particle at time t + velocity x (time taken to reach the arbitrary time from time t)

Therefore, the position vector of the particle at an arbitrary time is given as r = [tex]r_0[/tex] + vt where:

[tex]r_0[/tex] is the position vector of the particle at time t. v is the velocity of the particle. t is the time taken to reach the arbitrary time from time t.

For instance, if the particle passes through the origin at time t, moving with constant velocity v, the position vector of the particle at an arbitrary time will be given as;

r = 0 + vt = vt

Hence, the position vector of the particle at an arbitrary time is vt.

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The intensity of sound at the busy sound corner is 3.162 × 10⁻² W/m².

The sound intensity, represented by I, is defined as the power conveyed by a sound wave per unit area. Watts per square metre (W/m2) are the units of measurement.

waves are a type of energy propagation through a medium by means of adiabatic  lading and unloading. Important amounts for describing  aural  swells are  aural pressure,  flyspeck  haste,  flyspeck  relegation and  aural intensity.

The formula for determining sound intensity from decibel level is as follows:

I = I₀ × 10^(L/10)

where I0 is the reference intensity and L is the decibel level.

Plugging in the values from the issue yields:

I = (1×10⁻¹² W/m²) × 10^(75/10) = 3.162 × 10⁻² W/m²

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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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