Ask question

Ask question

All categories

3 years ago

11

If your car was traveling at 60 mph, that is about 27m/s. How long would it take a car traveling at 27m/s to slow to a stop, if

it slowed at the maximum safe rate of -8m/s2?

1 answer:

RUDIKE [14]3 years ago

6 0

Answer:

vf=at+v0. vf=0

-v0=at

t=-v0/a=(-27m/s)/(-8m/s^2)=3.4s

You might be interested in

If you were thinking about a washing machine as a system which of the following represents the inputs?

Lostsunrise [7]

Answer:

The answer is a, the dirty cloths, water and detergent.

Explanation:

The answer is the above selected because the inputs basically represent the data that are passed through the system to generate the output.

In this case, the inputs are the aforementioned in the answer while the possible output would literally be the clean cloths.

4 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago

A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.

Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

$PE=KE\\mgh = \frac{1}{2}mv^2$

where

m = 0.160 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

$v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s$

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

$s=(\frac{u+v}{2})t$

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

$t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s$

C)

The force exerted by the floor on the ball can be found using the equation:

$F=\frac{\Delta p}{t}$

where

$\Delta p$ is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

$\Delta p = m(v-u)$

So the equation can be rewritten as

$F=\frac{m(v-u)}{t}$

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

$F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N$

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0

3 years ago

Help plzzz!!! I only have 10 minutes to turn in

Lostsunrise [7]

The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car. When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

Kinetic energy = (1/2) (mass) (speed²)

And there we have it

The car's mass is 3,600 kg.
Its speed is 'v' m/s .
(1/2) (mass) (v²) = 5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

v² = (5406/1800) (Newton-meter/kg)

v² = (5406/1800) (kg-m²/s² / kg)

v² = (5406/1800) (m²/s²)

= = = = =

v = √(5406/1800) m/s

<em>v = 1.733 m/s</em>

4 0

3 years ago

A person's center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass mm at the top of a

muminat

Answer:

$\mathbf{v_{max} = \sqrt{gL}}$

Explanation:

Considering an object that moving about in a circular path, the equation for such centripetal force can be computed as:

$\mathbf {F = \dfrac{mv^2}{2}}$

The model for the person can be seen in the diagram attached below.

So, along the horizontal axis, the net force that is exerted on the person is:

$mg cos \theta = \dfrac{mv^2}{L}$

Dividing both sides by "m"; we have :

$g cos \theta = \dfrac{v^2}{L}$

Making "v" the subject of the formula: we have:

$v^2 = g Lcos \theta$

$v=\sqrt{ gL cos \theta$

So, when $\theta$ = 0; the velocity is maximum

∴

$v_{max} = \sqrt{gL \ cos \theta}$

$v_{max} = \sqrt{gL \ cos (0)}$

$v_{max} = \sqrt{gL \times 1}$

$\mathbf{v_{max} = \sqrt{gL}}$

Hence; the maximum walking speed for the person is $\mathbf{v_{max} = \sqrt{gL}}$

3 0

3 years ago