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finlep [7]
3 years ago
9

The radius of The moon is one_fourth and its mass is one eighty-first that of The earth. If The acceleration due to gravity on T

he surface of The earth is 9.8m/s square what is its value on The moon surface (pls ans that for me)...
Physics
1 answer:
77julia77 [94]3 years ago
7 0
g_{moon} =  g_{earth}*  \frac{( \frac{1}{81} )}{ (\frac{1}{ 4^{2}})}  = g_{earth} *  \frac{16}{81} = 9.8*16/81= 1.935
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What is vector quantities?
WINSTONCH [101]

Answer: Vector quantity is a measurement that has both magnitude and direction.

Explanation:

Vector quantity is a measurement that has both magnitude and direction. Examples are force, acceleration, displacement and velocity.

5 0
3 years ago
An object has a position given by r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s2)t2] j^, where all quantities are in SI units.
makkiz [27]

Answer:

Acceleration of the object is 4\ m/s^2.

Explanation:

It is given that, the position of the object is given by :

r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j

Velocity of the object, v=\dfrac{dr}{dt}

Acceleration of the object is given by :

a=\dfrac{d^2r}{dt^2}

a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)

Using the property of differentiation, we get :

a=\dfrac{d^2r}{dt^2}=-4\ m/s^2

So, the magnitude of the acceleration of the object at time t = 2.00 s is 4\ m/s^2. Hence, this is the required solution.

5 0
3 years ago
A 2.0-kg mass is oscillating about the origin at 24 rad/s. The amplitude of the oscillations is 0.040 m. At what position is the
Darya [45]

Answer:

0.0327 m

Explanation:

m = 2 kg

ω = 24 rad/s

A = 0.040 m

Let at position y, the potential energy is twice the kinetic energy.

The potential energy is given by

U = 1/2 m x ω² x y²

The kinetic energy is given by

K = 1/2 m x ω² x (A² - y²)

Equate both the energies as according to the question

1/2 m x ω² x y² = 2 x 1/2 m x ω² x (A² - y²)

y² = 2 A² - 2 y²

3y² = 2A²

y² = 2/3 A²

y = 0.82 A = 0.82 x 0.040 = 0.0327 m

4 0
3 years ago
A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 31.8°, the block starts
galina1969 [7]

Answer:

\mu_s = 0.62

\mu_k = 0.415

The motion of the block is downwards with acceleration 1.7 m/s^2.

Explanation:

First, we will calculate the acceleration using the kinematics equations. We will denote the direction along the incline as x-direction.

x - x_0 = v_0t + \frac{1}{2}at^2\\3.4 = 0 + \frac{1}{2}a(2)^2\\a = 1.7~m/s^2

Newton’s Second Law can be used to find the net force applied on the block in the -x-direction.

F = ma\\F = 1.7m

Now, let’s investigate the free-body diagram of the block.

Along the x-direction, there are two forces: The x-component of the block’s weight and the kinetic friction force. Therefore,

F = mg\sin(\theta) - \mu_k mg\cos(\theta)\\1.7m = mg\sin(31.8) - \mu_k mg\cos(31.8)\\1.7 = (9.8)\sin(\theta) - mu_k(9.8)\cos(\theta)\\mu_k = 0.415

As for the static friction, we will consider the angle 31.8, but just before the block starts the move.

mg\sin(31.8) = \mu_s mg\cos(31.8)\\\mu_s = tan(31.8) = 0.62

5 0
3 years ago
A rightward force is applied to a box in order to move it across the table at a constant velocity. (Ignore wind resistance).
anzhelika [568]
The forces would be gravity and friction
3 0
3 years ago
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