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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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The correct answer is D. Amount of time and area of physical contact between the substances.

Explanation:

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If 100 kg person weighed 400 N on the planet Zorg,
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Answer:

<u>4 m/s²</u>

Explanation:

<u>Formula</u>

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What is the name of the chart that contain elements?​
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A magnetic B field of strength 0.9 T is perpendicular to a loop with an area of 3 m2. If the area of the loop is reduced to zero
Alexeev081 [22]

Answer:

The magnitude of induced emf is 5.4 V

Explanation:

Given:

Magnetic field B = 0.9 T

Area of loop \Delta A = 3 m^{2}

Time take to reduce loop to zero \Delta t = 0.5 sec

To find induced emf we use faraday's law,

Induced emf is given by,

  \epsilon = -\frac{\Delta \phi}{\Delta t}

Here minus sign shows lenz law, for finding magnitude of emf  we ignore it.

Where \phi = B\Delta A

Put the value of flux and find induced emf,

   \epsilon = \frac{B\Delta A}{\Delta t}

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3 years ago
B) If the actual counterweight fitted to this boom was
Lostsunrise [7]

Answer:

50 N

4.2 N

Explanation:

i) The force needed to balance the boom is 2400 N.  If the weight of the counterbalance is 2350 N, then the downward force the park attendant must apply is 50 N.

ii) When the boom is resting on the end support, the normal force is:

∑τ = Iα

-W (0.50) + F (3.0) − N (6.0) = 0

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N = (-0.50 W + 3.0 F) / 6.0

N = (-0.50 × 2350 + 3.0 × 400) / 6.0

N ≈ 4.2

6 0
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