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antoniya [11.8K]

3 years ago

13

A speeding Thunderbird left skid marks on the road that were 76.7 m long when it came to a stop. If the acceleration of the car

was -10 m/s2, how fast was the car initially going?

1 answer:

FrozenT [24]3 years ago

4 0

Answer: 39.2 m/s

Explanation:

You can use the kinematic equation:

$v_f^2=v_i^2+2a*d$

We know the final velocity because it says it came to a stop. So now all we gotta do is plug in.

$0^2=v_i^2+2(-10)(76.7)\\v_i^2=1,534\\v_i=\sqrt{1534} \\=39.166 m/s$

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A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is currently supplying 150N of force.

alexdok [17]

Here when an object is placed on the level floor then in that case there are two forces on the object

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so when no force is applied on the box at that time normal force is counter balanced by weight.

Now here it is given that A person tried to lift the box upwards

So now there are two forces on the box

1). Applied force of person

2). Normal force due to ground

So now these two forces will counter balance the weight of the crate

So we can write an equation for force balance like

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given that

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here

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$F_n = 150 N$

now from above equation

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7 0

3 years ago

A kangaroo can jump straight up to a height of 2.0 m. What is its takeoff speed

12345 [234]

7.17m/s glad I could help

5 0

3 years ago

A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They

soldier1979 [14.2K]

Answer:

a. $\thta=71^{\circ}$

b. $v_f=3.78 m/s$

Explanation:

We are given that

$m_1=53 kg$

$v_1=2.9 m/s$

$m_2= 72 kg$

$v_2=6.2 m/s$

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$\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}$

$\theta=71^{\circ}$

b. We have to find the speed $v_f$

According to law of conservation of momentum

$m_1v_1=(m_1+m_2)v_fcos\theta$

$53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f$

$v_f=\frac{153.7}{40.7}=3.78 m/s$

3 0

3 years ago

Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter

valentinak56 [21]

Wee can use here kinematics

as we know that

$y = v*t + \frac{1}{2} at^2$

for shorter tree we know that

$y = 0 + \frac{1}{2}*9.8 * 2^2$

$y = 19.6 meter$

now since we know that other tree is twice high

So height of other tree is y = 39.2 m

now again by above equation

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$39.2 = 0 + \frac{1}{2}*9.8 * t^2$

$t = 2.83 s$

so the time taken is 2.83 s

4 0

3 years ago