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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water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. the density of water is 1 000 kg/m3. determine its average velocity. multiple choice question. 20 m/s 200 m/s 0.02 m/s 2 m/s 0.2 m/s
Option D: 2 m/s is the average velocity of the water flowing through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s.
According to the question:
cross-sectional area of the pipe = 0.002m²
Mass flowrate = 4 kg/s
Density of water = 1000 kg/m³
We are asked to find, average velocity =?
Average velocity is the net or total displacement covered by a body in a given time. The mass flow rate divided by the pipe's cross-sectional area and density ratio is the formula for calculating a fluid's average velocity.
As a result, the water's average flow rate through the pipe is provided by:
v = m / (ρ × A)
where, v is the average velocity, m is the mass flow rate, ρ is the density of water, and A is the cross-sectional area of the pipe. Substituting the values in the above equation we get:
v = 4 / (1000 × 0.002)
v = 2m/s
Therefore, the average velocity of water flowing through a pipe of cross-sectional area of 0.002m² is 2m/s.
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Correct question is:
Water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. The density of water is 1 000 kg/m3. Determine its average velocity. Multiple choice question.
20 m/s
200 m/s
0.02 m/s
2 m/s
0.2 m/s
a 10.0-mf capacitor is fully charged across a 12.0-v bat- tery. the capacitor is then disconnected from the battery and connected across an initially uncharged capacitor with capacitance c. the resulting voltage across each capacitor is 3.00 v. what is the value of c?
The value of uncharged capacitor in series with a 10.0-microfarad capacitor, given that each capacitor has a voltage of 3.00 volts, can be calculated using the formula for equivalent capacitance in series and formula for charge on a capacitor. The value of c is approximately 4.00 microfarads.
To determine the value of c, which is [tex]1/Ceq = 1/C1 + 1/C2[/tex] . Initially, the 10.0-microfarad capacitor has a charge of [tex]Q = CV = (10.0 * 10^{-6 }F) * 12.0 V = 1.20 * 10^{-4} C[/tex].
When it is connected in series with uncharged capacitor with capacitance c, charge is shared between the two capacitors. The charge on 10.0-microfarad capacitor is also equal to the charge on uncharged capacitor, which is given by [tex]Q = (3.00 V) * C[/tex] .
Equating the two expressions for Q and solving for c, we get [tex]C = Q/3.00[/tex] [tex]V = (1.20 * 10^{-4 C}) / (3.00 V) = 4.00 * 10^{-5 F}[/tex]. Therefore, value of c is approximately 4.00 microfarads.
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a mass-spring oscillating system undergoes shm with a period t. what is the period of the system if the amplitude is doubled?
The period of a mass-spring oscillating system undergoing SHM with a period t, when the amplitude is doubled, is still t.
The period of a mass-spring oscillating system undergoing simple harmonic motion (SHM) is determined by the spring constant and mass of the system.
When the amplitude of the system is doubled, the period of the system remains the same, regardless of the amplitude. This means that the period of a mass-spring oscillating system undergoing SHM with a period t, when the amplitude is doubled, is still t.
To understand why the period remains the same, consider the equation for simple harmonic motion:
x(t) = A cos (2πft).
This equation describes the displacement of an object over time and is based on the principle that any system undergoing SHM oscillates about a fixed point at a constant frequency.
The frequency of the system is inversely proportional to the period, and is determined by the spring constant and mass of the system.
Increasing the amplitude of the system does not affect the frequency or period of the oscillations.
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Why is momentum not conserved in real life situations
Momentum is not always conserved in real-life situations because external forces can act on a system and change its momentum.
For example, when two cars collide, friction and air resistance can cause the momentum of the system to change. Similarly, when a ball is thrown in the air, gravity and air resistance act on it and cause its momentum to change. Other factors such as deformation, energy loss, and imperfect collisions can also cause momentum to be lost or gained. Therefore, while momentum is a useful concept in physics, it is important to consider the impact of external factors when analyzing real-world situations.
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two stationary point charges q1 and q2 are shown in the figure along with a sketch of some field linesrepresenting the electric field produced by them. what can you deduce from the sketch?
From the sketch, we can deduce that the two charges q1 and q2 are of opposite signs, as field lines start at the positive charge q1 and end at the negative charge q2. The field lines also indicate that the magnitude of the electric field produced by q1 is larger than that of q2.
Additionally, the field lines show that the electric field lines near the charges are denser, indicating a stronger electric field intensity near the charges. The direction of the electric field points from q1 to q2, which is consistent with the direction of the force that a positive test charge would experience if placed in the field. The field lines also show that the electric field is radial, i.e., the field lines point directly away from or towards each charge in a straight line, which is a characteristic of the electric field produced by a point charge. Finally, the density of the field lines decreases with distance from the charges, indicating that the electric field strength decreases with distance from the charges, following an inverse-square law.Learn more about electric field at: https://brainly.com/question/14372859
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a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet. the coefficients of static and kinetic friction are 0.6 and 0.4, respectively. the worker pushes with a force of 600 n. the frictional force exerted by the surface is
When a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet, the frictional force exerted by the surface on the crate is 588 N.
When a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet, the frictional force exerted by the surface can be calculated as follows:
The weight of the crate = m × g = 100 kg × 9.8 m/s² = 980 N
Force applied by the worker = F = 600 N
The force of friction acting on the crate is given by the following formula:
Ff = μF
Where, μ is the coefficient of friction, F is the normal force acting on the crate.
Notes: The normal force is equal and opposite to the weight of the crate. i.e., N = 980 N1. The frictional force exerted by the surface on the crate is the static frictional force initially. Hence, we use the coefficient of static friction for our calculation.
2. If the force applied by the worker is not enough to overcome the static frictional force, then the crate will not move and the frictional force will remain static friction.
3. Once the crate starts moving, the static friction will convert to kinetic friction. Hence, we will use the coefficient of kinetic friction if the force applied by the worker is greater than the force of static friction. Initially, the force applied by the worker is less than the force of static friction, hence the frictional force exerted on the crate will be the static frictional force.
Frictional force = Ff = μN
The normal force acting on the crate = Weight of the crate = 980 N
Frictional force =
Ff = μN
= 0.6 × 980 N
= 588 N
Therefore, the frictional force exerted by the surface on the crate is 588 N.
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