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Alisiya [41]
3 years ago
13

A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves 1.5 m before it comes to a complete stop. If the

car had been moving at 14 m/s, how far would it have continued to move after the brakes were applied?
a. 1.5m
b. 3.0 m
c. 4.5 m
d. 6.0 m
e. 7.5 m
Physics
1 answer:
viva [34]3 years ago
4 0

Answer:

d. 6.0 m

Explanation:

Given;

initial velocity of the car, u = 7.0 m/s

distance traveled by the car, d = 1.5 m

Assuming the car to be decelerating at a constant rate when the brakes were applied;

v² = u² + 2(-a)s

v² = u² - 2as

where;

v is the final velocity of the car when it stops

0 = u² - 2as

2as = u²

a = u² / 2s

a = (7)² / (2 x 1.5)

a = 16.333 m/s

When the velocity is 14 m/s

v² = u² - 2as

0 = u² - 2as

2as = u²

s = u² / 2a

s = (14)² / (2 x 16.333)

s = 6.0 m

Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.

The correct option is d

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AnnyKZ [126]
Hey there!

Here is your answer:

<u><em>The proper answer to this question is option C "</em></u><span><u><em>0.00349".</em></u>

Reason:

</span><span><u><em>1 L = 100 cL. Or 1 cL = 0.01 L</em></u>

</span><span><u><em>34.9 cL = 34.9 / 100 L = 0.349 L</em></u>

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<em>Therefore the answer is option C!</em>

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit</span>
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3 years ago
Time dilation: A missile moves with speed 6.5-10 m/s with respect to an observer on the ground. How long will it take the missil
tatyana61 [14]

Answer:

The time taken by missile's clock is 4.6\times 10^{6} s

Solution:

As per the question:

Speed of the missile, v_{m = 6.5\times 10^{3}} m/s

Now,

If 'T' be the time of the frame at rest then the dilated time as per the question is given as:

T' = T + 1

Now, using the time dilation eqn:

T' = \frac{T}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T'}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

\frac{T + 1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = \frac{1}{\sqrt{1 + (\frac{v_{m}}{c})^{2}}}

1 + \frac{1}{T} = (1 + (\frac{v_{m}}{c})^{2})^{- \frac{1}{2}}         (1)

Using binomial theorem in the above eqn:

We know that:

(1 + x)^{a} = 1 + ax

Thus eqn (1) becomes:

1 + \frac{1}{T} = 1 - \frac{- 1}{2}.\frac{v_{m}^{2}}{c^{2}}

T = \frac{2c^{2}}{v_{m}^{2}}

Now, putting appropriate values in the above eqn:

T = \frac{2(3\times 10^{8})^{2}}{(6.5\times 10^{3})^{2}}

T = 4.6\times 10^{6} s

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3 years ago
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Vlad1618 [11]

Answer:

a counterclaim

Explanation:

authors purpose is what an author wrote somthing for

opinion is someones thoughts or "side" on a argument

an arguement is a battle of opinions if that makes sense

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gladu [14]

Answer:

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Explanation:

v = speed of the proton = 8 x 10⁶ m/s

B = magnitude of the magnetic field = 1.72 T

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N

θ = Angle between proton's velocity and magnetic field

magnitude of magnetic force on the proton is given as

F = q v B Sinθ

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Sinθ = 0.327

θ = 19.1 deg

4 0
3 years ago
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