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3 years ago

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In severe head-on automobile accidents, a deceleration of 60 g’s or more (1 g 5 32.2 ft/s2) often results in a fatality. What fo

rce, in lbf, acts on a child whose mass is 50 lb, when subjected to a deceleration of 60 g’s?

1 answer:

Musya8 [376]3 years ago

3 0

Answer:

The resulting force on the child is 3000 lbf

Explanation:

To find the force that acts on a child of 50 lb with a deceleration of 60 g's, we can use the formula:

Force = mass * acceleration

To find the force in lbf, we need to use the mass in lb and the acceleration in g (standard unit of gravity).

So we have that:

Force = 50 * 60

Force = 3000 lbf

So the resulting force on the child is 3000 lbf.

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3 years ago

A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin

pychu [463]

a) See graph in attachment

b) The suvat equation to use is $v_f^2 - v_i^2 = 2as$

c) The acceleration is $a=\frac{v_f^2-v_i^2}{2s}$

d) The acceleration is $1.25 m/s^2$

e) The time needed is 8 s

Explanation:

a)

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is $x_i = 0$ , the origin

- The final position of the boat is $x_f = 200 m$

- The initial velocity of the boat is $v_i = 20.0 m/s$

- The final velocity of the boat is $v_f = 30.0 m/s$

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

b)

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

$v_f^2 - v_i^2 = 2as$

where

$a$ is the acceleration

c)

Now we have to solve the equation

$v_f^2 - v_i^2 = 2as$

In order to find the acceleration.

This can be done by dividing both terms by $2s$ : this way, we find

$\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}$

And so the acceleration is

$a=\frac{v_f^2-v_i^2}{2s}$

d)

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

$v_i = 20.0 m/s$ , the initial velocity

$v_f = 30.0 m/s$ , the final velocity

$s = x_f - x_i = 200 m$ , the displacement of the boat

And substituting into the equation,

$a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2$

e)

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

$v_f = v_i + at$

where:

$v_i$ is the initial velocity

$v_f$ is the final velocity

a is the acceleration

t is the time

Here we have:

$v_i = 20.0 m/s$

$v_f = 30.0 m/s$

$a=1.25 m/s^2$

Solving for t, we find:

$t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s$

Learn more about accelerated motion:

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3 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Veseljchak [2.6K]

Answer:

The speed of q₂ is $4\sqrt{10}\ m/s$

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

$E_{i}=E_{f}$

$\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2$

$\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})$

Put the value into the formula

$\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})$

$0.00075(400-v_{f}^2)=0.18
$

$400-v_{f}^2=\dfrac{0.18}{0.00075}$

$-v_{f}^2=240-400$

$v_{f}^2=160$

$v_{f}=4\sqrt{10}\ m/s$

Hence, The speed of q₂ is $4\sqrt{10}\ m/s$

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