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3 years ago

What average net force is required to stop a 1950 kg car in 10.5 s if it’s initially traveling at 28m/s

Physics

1 answer:

nika2105 [10]3 years ago

7 0

Answer:

An average net force of 5200 N is needed to stop the car

Explanation:

Cinematics and Dynamics

Cinematics describes the variables involved in the movement without dealing with its causes. There are four main concepts in cinematics: Velocity (or its scalar equivalent, the speed), acceleration, time, and displacement (or the scalar equivalent, distance).

The acceleration can be calculated by:

$\displaystyle a=\frac{v_f-v_o}{t}$

The initial speed is vo=28 m/s, it stops (vf=0) in t=10.5 seconds, thus the acceleration is:

$\displaystyle a=\frac{0-28}{10.5}$

$a = -2.67~m/s^2$

The acceleration is negative because the car loses speed.

Knowing the mass of the car m=1950 Kg, we can calculate the net force required to stop the car by using the formula:

F = m.a =1950*2.67

F = 5200 N

An average net force of 5200 N is needed to stop the car

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Answer:

The final temperature of the gas is 114.53°C.

Explanation:

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3 years ago

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

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n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

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$P=5\times 0.148$

P=740 KPa

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$\Delta =0.148\times 50$

Δ=7.4 mm

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3 years ago

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Hello,

Here is your answer:

The proper answer to this question is "because of there substantial size the rock rests on another rock which keeps it balanced".

If you need anymore help feel free to ask me!

Hope this helps!

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