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mylen [45]
3 years ago
6

SOMEONE PLSSSS HELP ME WITH THIS QUESTION!!

Physics
1 answer:
Hitman42 [59]3 years ago
4 0
The answer is indeed B
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The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/hr to 55 km.hr. the tires have a diamet
sveta [45]
<span>95 km/h = 26.39 m/s (95000m/3600 secs) 55 km/h = 15.28 m/s (55000m/3600 secs) 75 revolutions = 75 x 2pi = 471.23 radians radius = 0.80/2 = 0.40m v/r = omega (rad/s) 26.39/0.40 = 65.97 rad/s 15.28/0.40 = 38.20 rad/s s/((vi + vf)/2) = t 471.23 /((65.97 + 38.20)/2) = 9.04 secs (vf - vi)/t = a (38.20 - 65.97)/9.04 = -3.0719 The angular acceleration of the tires = -3.0719 rad/s^2 Time is required for it to stop (0 - 38.20)/ -3.0719 = 12.43 secs How far does it go? 65.97 - 38.20 = 27.77 M</span>
7 0
3 years ago
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 seamount is an isolated land mass rising from the ocean floor.<br> a. True<br> b. False
iris [78.8K]
The answer is A, True.
7 0
3 years ago
2. Tony sets up an experiment inside his classroom. He spaces 5 plants at equal
brilliants [131]

Answer:

a

Explanation:

4 0
4 years ago
Read 2 more answers
Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0
3 years ago
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