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2 years ago

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WallyGPX accelerates from 0 m/s to 8 m/s in 3 seconds. What is his acceleration? Is this acceleration higher than that of a car

which accelerates from 0 to 30 m/s in 8 seconds?

1 answer:

olga nikolaevna [1]2 years ago

4 0

My Phone is +2348181686682

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When driving in heavy rain, or on a flooded road, your tires can ride on a thin film of water like skis;

Simora [160]

The answer is letter a. It is best to slow down in situations of heavy rain or flooded road as skid could be the result if you lose out of control because the driver isn't slowing down. That is why it is being said that tires can ride on a thin film of water skis as it could skid if it has lost control if the driver hadn't slowed down.

7 0

1 year ago

A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of

olya-2409 [2.1K]

Explanation:

When Michelson-Morley apparatus is turned through $90^{o}$ then position of two mirrors will be changed. The resultant path difference will be as follows.

$\frac{lv^{2}}{\lambda c^{2}} - (-\frac{lv^{2}}{\lambda c^{2}}) = \frac{2lv^{2}}{\lambda c^{2}}$

Formula for change in fringe shift is as follows.

n = $\frac{2lv^{2}}{\lambda c^{2}}$

$v^{2} = \frac{n \lambda c^{2}}{2l}$

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.

l = 11 m

$\lambda = 5.9 \times 10^{-7} m$

c = $3.0 \times 10^{8}$ m/s

Hence, putting the given values into the above formula as follows.

v = $\sqrt{\frac{n \lambda c^{2}}{2l}}$

= $\sqrt{\frac{1 \times (5.9 \times 10^{-7} m) \times (3.0 \times 10^{8})^{2}}{2 \times 11 m}}$

= $2.41363 \times 10^{9} m/s$

Thus, we can conclude that velocity deduced is $2.41363 \times 10^{9} m/s$ .

3 0

1 year ago

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

<u>Answer:</u>

Cannonball will be in flight before it hits the ground for 2.02 seconds

<u>Explanation:</u>

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

5 0

1 year ago

A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0Â° east of

koban [17]

<u>Answer:</u>

Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

Let the east point towards positive X-axis and north point towards positive Y-axis.

Speed of truck = 25 m/s north = 25 j m/s

Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

= (1.43 i + 1.00 j) m/s

Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

Magnitude of velocity = 26.04 m/s

Angle from positive horizontal axis = 86.85⁰

So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0

1 year ago

The chart shows data for four moving objects. A 4 column table with 4 rows. The first column is labeled Object with entries, W,

KatRina [158]

Answer:

y

Explanation:

I took the test

3 0

1 year ago