Ask question

Ask question

All categories

2 years ago

9

A runner whose mass is 54 kg accelerates from a stop to a speed of 7 m/s in 3 seconds. (A good sprinter can run 100 meters in ab

out 10 seconds, with an average speed of 10 m/s.) (a) What is the average horizontal component of the force that the ground exerts on the runner's shoes? (b) How much work is done on the point-particle system by this force?

1 answer:

Darya [45]2 years ago

6 0

Answer:

a. $F=126N$

b. $E_K=1323J$

Explanation:

Given:

$m=54kg$

$v=7 m/s$

$t= 3s$

The runner force average to find given the equations

a.

$F=m*a$

$a=\frac{v}{t}$

$F=m*\frac{v}{t}=54kg*\frac{7m/s}{3s}$

$F=126N$

b.

Work done by the system by this force so

$W=F*d$

$W=E_K$

$E_K=\frac{1}{2}*m*v^2$

$E_K=\frac{1}{2}*54kg*(7m/s)^2$

$E_K=1323J$

You might be interested in

On a velocity vs. time graph, what does it mean when the line crosses over the x-axis?

Phantasy [73]

On a velocity - time graph, if the line crosses the x - axis it depicts that the object has started moving in the opposite direction.

8 0

3 years ago

Please helppppppp!!!!!!!!!!!!!!

azamat

Answer:

circuit breaker

Explanation:

A circuit breaker is a device used for electrical safety. It consists of a switch designed to protect an electrical circuit from damage that may result from heating due to overload in the circuit.

Its basic function is to interrupt current flow through its switch that consists of metal stripe which bends when it gets hot.

Fuse has similar action with circuit breaker, the only difference is that fuse can only be used once because it melts when it gets hot.

Therefore, the correct answer is "circuit breaker"

8 0

3 years ago

If a fish is trying to capture an insect hovering above the surface of the water – how will it jump to catch it? Will it aim abo

Viktor [21]

Answer:

It will have aim at a point "below" the insect.

From the insect's point of view, the fish will appear to be shallower than it actually is because a ray of light from the insect to the fish will be bent "towards" the normal when the ray enters the water

5 0

2 years ago

What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of?

madreJ [45]

We will apply the Newton's second Law so the we will be able to find the acceleration.
F (tot) = ma
a = F(tot) / m
a = 32.0 N / 65.0 kg = 0.492 m/s^2
Approximately 0.492 m/s^2 is her initial acceleration if she is initially stationary and wearing steel-bladed skates.

7 0

3 years ago

A mass m1 hangs from a spring k and is in static equilibrium. A second mass m2 drops through a height h and sticks to m1 without

jekas [21]

Answer:

$\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}$

Explanation:

From the information given:

The equation of the motion can be represented as:

$(m_1 +m_2) \hat u + ku = m_2 g--- (1)$

where:

$m_1$ = mass of the body 1

$m_2$ = mass of the body 2

$\hat u$ = acceleration

k = spring constant

u = displacement

g = acceleration due to gravity

Recall that the formula for natural frequency $\omega _n = \sqrt{\dfrac{k}{m_1+m_2}}$

And the equation for the general solution can be represented as:

$u(t) = A cos \omega_nt + B sin \omega _n + \dfrac{m_2g}{k} --- (2)$

To determine the initial velocity, we have:

$\hat u_2^2 = 2gh$

$\hat u_2 = \sqrt{2gh}$

where h = height

Suppose we differentiate equation (2) with respect to time t; we have the following illustration:

$\hat u (t) = - \omega_n A sin \omega_n t+ \omega_n B cos \omega _n t + 0$

now if t = 0

Then

$\hat u (0) = - \omega_n A sin \omega_n (0)+ \omega_n B cos \omega _n (0) + 0$

$= \omega _n B$

Using the law of conservation of momentum on the impact;

$m_2 \hat u_2=(m_1+m_2) \hat u (0)$

By replacing the value of $\hat u_2$ with $\sqrt{2gh$

Then the above equation becomes:

$m_2 \times \sqrt{2gh}=(m_1+m_2) \ u(0)$

Making u(0) the subject of the formula, we have:

$u(0)= \dfrac{ m_2 \times \sqrt{2gh}}{(m_1+m_2)}$

Similarly, the value of the variable can be determined as follows;

Using boundary conditions

$0 = A cos 0 + B sin 0 + \dfrac{m_2g}{k}$

$0 = A (1)+0+ \dfrac{m_2g}{k}$

$A =- \dfrac{m_2g}{k}$

Also, if $\hat u (0) = \omega_nB$

Then :

$\dfrac{m_2}{m_1+m_2}\sqrt{2gh} = \omega_n B$

making B the subject; we have:

$B = \dfrac{m_2}{m_1 + m_2}\dfrac{\sqrt{2gh}}{\omega_n}$

Finally, replacing the value of A and B back to the general solution at equation (2); we have the equation of the subsequent motion u(t) which is:

$\mathbf{u(t) =\dfrac{m_2g }{k}(1 - cos \omega_n t) + \dfrac{\sqrt{2gh}}{\omega_n}\times \dfrac{m_1}{m_1+m_2}sin \omega _n t}$

8 0

3 years ago