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Agata [3.3K]
3 years ago
10

A man claims that he can hold onto a 16.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man

in a collision in which he is in one of two identical cars each traveling toward the other at 59.0 mi/h relative to the ground. The car in which he rides is brought to rest in 0.05 s.
Find the magnitude of the average force needed to hold onto the child.
N
Physics
1 answer:
Anna007 [38]3 years ago
5 0

Answer:

F = -8440.12 N

the magnitude of the average force needed to hold onto the child is 8440.12 N

Explanation:

Given;

Mass of child m = 16 kg

Speed of each car v = 59.0 mi/h = 26.37536 m/s

Time t = 0.05s

Applying the impulse momentum equation;

Impulse = change in momentum

Ft = ∆(mv)

F = ∆(mv)/t

F = m(∆v)/t

Where;

F = force

t = time

m = mass

v = velocity

Since the final speed of the car is zero(at rest) then;

∆v = 0 - v = -26.37536 m/s

Substituting the given values;

F = 16×-26.37536/0.05

F = -8440.1152 N

F = -8440.12 N

the magnitude of the average force needed to hold onto the child is 8440.12 N

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Two rocks are at the top of a building. Rock 1 is dropped from rest while Rock 2 is thrown horizontaly at a velocity of 5 ms.
LuckyWell [14K]

Answer:

I'm pretty sure the answer is 0 m/s²

Explanation:

The horizontal velocity of the second rock is 5 m/s, so if we pretend air resistance doesn't exist, it will maintain that horizontal velocity, meaning that there is no horizontal acceleration.

7 0
3 years ago
What type of cooling would a scientist determine formed an igneous rock found with large crystals? Slow cooling Medium rate cool
Dmitriy789 [7]

Slower cooling engenders the growth of larger crystals in igneous rocks, thus, your answer should be slow cooling!

Hope this helped!

5 0
3 years ago
Read 2 more answers
The device used to measure a masses of a body is kilogram . true or false​
olga nikolaevna [1]

Answer: false

Explanation:

While kilograms are the unit used to measure body mass, the device used is a scale.

Hope it helps :)

6 0
3 years ago
Water near the poles would most likely be stored as
solniwko [45]
Hey there! 

Answer: Glaciers

Water near the poles would most likely be stored as glaciers. Glaciers are slow moving rivers that are a buildup of ice and snow. 

Thank you!

5 0
3 years ago
A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge
zheka24 [161]

Answer:

The acceleration of an electron is 1.2\times10^{20}\ m/s^2

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

F_{1}=\dfrac{kq_{1}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}

F_{1}=8\times10^{-11}j\ \ N

We need to calculate the force on electron due to q₂

Using formula of force

F_{2}=\dfrac{kq_{2}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}

F_{2}=6.93\times10^{-11}i\ \ N

We need to calculate the net force

Using formula of net force

F=F_{1}+F_{2}

Put the value into the formula

F=8\times10^{-11}j+6.93\times10^{-11}i

The magnitude of the net force

F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}

F=1.058\times10^{-10}\ N

We need to calculate the acceleration of an electron

Using formula of force

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}

a=1.2\times10^{20}\ m/s^2

Hence, The acceleration of an electron is 1.2\times10^{20}\ m/s^2

3 0
3 years ago
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