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Ray Of Light [21]
3 years ago
9

Which heavenly bodies are involved in causing the earths tides?

Physics
1 answer:
Elis [28]3 years ago
4 0
All of them except any planet means "Earth, Moon and the Sun"

so, option C is your answer.

Hope this helps!
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An object is moving in a straight line along the y axis. As a function of time, its position is given by the equation y=3.0+4.8x
romanna [79]

Answer:

Explanation:

<u>Instant Velocity and Acceleration </u>

Give the position of an object as a function of time y(x), the instant velocity can be obtained by

v(x)=y'(x)

Where y'(x) is the first derivative of y respect to time x. The instant acceleration is given by

a(x)=v'(x)=y''(x)

We are given the function for y

y(x)=3.0+4.8x +6.4x^2

Note we have changed the last term to be quadratic, so the question has more sense.

The velocity is

v(x)=y'(x)=4.8+12.8x

And the acceleration is a(x)=v'(x)=12.8

5 0
3 years ago
A football player kicks a ball with a mass of 0.42 kg. The average acceleration of the football was 14.8 m/s².
kvasek [131]
D.6.22N. because .42kg * 14.8m/s=6.22 N[meaning newtons}.
6 0
3 years ago
Read 2 more answers
Diamonds are usually found in pipes 50 to 200 m across made of ________.
Leokris [45]
I believe the correct answer would be kimberlite. Diamonds are usually found in pipes 50 to 200 m across made of kimberlite. It is an igneous rock that is known to contain traces of diamonds. It is named base on the town where it was discovered which is Kimberley, South Africa. 



7 0
3 years ago
A non-_____ rock has interlocking grains with no specific pattern.
Kazeer [188]
A non <span>foliated </span>rock has interlocking grains with no specific pattern.
3 0
3 years ago
Read 2 more answers
A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

\omega=-\alpha t

\alpha=-\dfrac{\omega}{t}

\alpha=-\dfrac{50.0}{20.0}

\alpha=-2.5\ rad/s^2

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

\theta=\omega_{0}t+\dfrac{1}{2}\alpha t

Put the value into the formula

\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2

\theta=500\ rad

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

\vec{\tau}=\vec{r}\times\vec{f}

\tau=r\times f\sin\theta

Put the value into the formula

\tau=2.5\times4\times 9.8\sin60

\tau=84.87\ N-m

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0
3 years ago
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