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vivado [14]
2 years ago
7

A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has

Physics
1 answer:
Helen [10]2 years ago
3 0

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

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During which phase of the moon do neap tides occur?
Fynjy0 [20]

Answer:

First Quarter and Third Quarter.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.

Since gravity variates with the distance:

F = G\frac{m1\cdot m2}{r^{2}} (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.

When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).

               

However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.

           

Therefore, that happens when the Moon is at First Quarter and Third Quarter.

4 0
3 years ago
Read 2 more answers
A student drops an egg from the roof of their house onto a trampoline. The egg feels a change in momentum of 2.2 kg^ * m/s in 1.
geniusboy [140]

Answer:

F = m a = m v / t       where v is the change in velocity in time t

F = p / t       since m v is equal to p

F = 2.2 (kg m / s) / 1.1 s = 2 kg-m / s^2 = 2 N

Or you can use the impulse equation

8 0
2 years ago
A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the
Marrrta [24]

Our values can be defined like this,

m = 65kg

v = 3.5m / s

d = 0.55m

The problem can be solved for part A, through the Work Theorem that says the following,

W = \Delta KE

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

\frac {1} {2} mv ^ 2

So,

Fd = \frac {1} {2} m ^ 2

Clearing F,

F = \frac {mv ^ 2} {2d}

Replacing the values

F = \frac {(65) (3.5)} {2 * 0.55}

F = 723.9N

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0
2 years ago
What is acceleration produced by a force of 12 newton exerted on an object of mass 3kg​
jarptica [38.1K]

Answer:

a=F/m

a=12N/3kg  (here newton can be written as kgm/s^2 so kg will be cancelled)

a=4m/s^2

Explanation:

3 0
3 years ago
A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping. His resultant displacement was
Serhud [2]

<u>Answer:</u>

 Resultant displacement is 4 km north.

<u>Explanation:</u>

    Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

   A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping,

   So initial displacement, 15 kilometers north = 15 j km

        Second displacement, 11 kilometers south = -11 j km

   Total displacement = 15 j -11 j = 4 j km

   Total displacement is 4 km north.

4 0
3 years ago
Read 2 more answers
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