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Sergeeva-Olga [200]

2 years ago

12

A 55 kg person is in a head-on collision. The car's speed at impact is 12 m/s. Estimate the net force on the person if he or she

is wearing a seat belt and if the air bag deploys.

1 answer:

ollegr [7]2 years ago

4 0

1.244 m per second the person driving will go

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is it possible to breed a black haired long tailed retriever? If so, what percentage is possible? PLEASE answer the corrects ans

Pachacha [2.7K]

Answer: it can be considered a genetic mutation with a history of a Golden Retriever in their blood but it is very rare. and there our some black retrievers you can buy too. i hope i helped

Explanation:

6 0

1 year ago

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which statement is a Hypothesis? A: Most of the earthworms moved to the shaded area during experiment. B: If an earthworm is giv

True [87]

Answer:

B

Explanation:

It's saying what you think

3 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

(c) $x_f=2.6m$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

(c) Using the kinematics equation:

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago

How long would it take an object to reach the ground from the top of a building that is 470 feet tall? Round to the nearest tent

Zinaida [17]

Answer:

It would take the object 5.4 s to reach the ground.

Explanation:

Hi there!

The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:

h = h0 + v0 · t + 1/2 · g · t²

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).

t = time

Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:

h = h0 + 1/2 · g · t²

We have to find the time at which h = 0:

0 = 470 ft - 1/2 · 32.2 ft/s² · t²

Solving for t:

-470 ft = -16.1 ft/s² · t²

-470 ft / -16.1 ft/s² = t²

t = 5.4 s

3 0

3 years ago

Larry lightweight stands on a pair of bathroom scales each scale reads 300 N what is larry's mass

leonid [27]

Larry's mass is <span>30.6 kg.</span>

7 0

3 years ago

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