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2 years ago

A high school physics student is sitting in a seat read-

1 answer:

7 0

The equilibrium condition allows finding the result for the force that the chair exerts on the student is:

The reaction force that the chair exerts on the student's support is equal to the student's weight.

Newton's second law gives the relationship between force, mass and acceleration of bodies, in the special case that the acceleration is is zero equilibrium condition.

∑ F = 0

Where F is the external force.

The free body diagram is a diagram of the forces on bodies without the details of the shape of the body, in the attached we can see a diagram of the forces.

Let's analyze the force on the chair.

$N_{chair} - W_{chair} - W_{student} = 0 \\ \\N_{chair} = W_{chair} + W_{student}$

Let's analyze the forces on the student.

$N_{student} - W_{student} = 0 \\N_{student} = W _{student}$

In conclusion using the equilibrium condition we can find the result for the force that the chair exerts on the student is:

The reaction force that the chair exerts on the student's support is equal to the student's weight.

Learn more here: brainly.com/question/18117041

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NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

3 years ago

Compared to the weak nuclear force the electromagnetic force

Crazy boy [7]

Explanation:

Weak nuclear force:

The interaction between the subatomic particles is called weak nuclear force.

The weak nuclear force is one of the four fundamental forces.

The weak nuclear force is effective at very short distance.

The range and relative strength of weak nuclear force is 10⁻¹⁸ m and 10²⁵ with respect to gravitational force respectively

Deuterium is formed due to the fusion of protons and neutrons under the action the weak force.

Example : Beta decay

Electromagnetic force:

The interaction between the charged particles is called electromagnetic force.

The electromagnetic force is one of the four fundamental forces.

The electromagnetic force is effective at long range distance.

The range and relative strength of electromagnetic force is infinity and 10³⁶ with respect to gravitational force respectively

Example : light

8 0

3 years ago

Read 2 more answers

Maggie participated in a race. She ran 100 meters west and then turned back and ran to the starting line to complete the race. M

earnstyle [38]

Her total distance was 200 meters, since she ran 100 meters west and the same 100 meters back. This gives an average speed of 200 m / 40 s = 5 m/s.

However, her displacement was 0, since she ended at the same place she began, therefore, her average velocity is 0 / 40 s = 0.

3 0

3 years ago

1. You released a pendulum of mass 1kg from a height of 0.05m

photoshop1234 [79]

a. The speed of the pendulum when it reaches the bottom is 0.9 m/s.

b. The height reached by the pendulum is 0.038 m.

c. When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

<h3>Kinetic energy of the pendulum when it reaches bottom</h3>

K.E = 100%P.E - 18%P.E

where;

P.E is potential; energy

K.E(bottom) = 0.82P.E

K.E(bottom) = 0.82(mgh)

K.E(bottom) = 0.82(1 x 9.8 x 0.05) = 0.402 J

<h3>Speed of the pendulum</h3>

K.E = ¹/₂mv²

2K.E = mv²

v² = (2K.E)/m

v² = (2 x 0.402)/1

v² = 0.804

v = √0.804

v = 0.9 m/s

<h3>Final potential energy </h3>

P.E = 100%K.E - 7%K.E

P.E = 93%K.E

P.E = 0.93(0.402 J)

P.E = 0.374 J

<h3>Height reached by the pendulum</h3>

P.E = mgh

h = P.E/mg

h = (0.374)/(1 x 9.8)

h = 0.038 m

<h3>when the pendulum stops</h3>

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

Thus, the speed of the pendulum when it reaches the bottom is 0.9 m/s.

The height reached by the pendulum is 0.038 m.

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

Learn more about pendulum here: brainly.com/question/26449711
#SPJ1

5 0

2 years ago

Three charges are arranged on the circle as shown in the figure Q1 = Q2 = 20μC. and Q3 = -10 μC on the y-axis. The electric fiel

Gnoma [55]

Explanation:

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3 years ago