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Evgesh-ka [11]
3 years ago
14

Compared to its mass on Earth, the mass of a 60-kg object on the moon is

Physics
1 answer:
vagabundo [1.1K]3 years ago
3 0

Compared to its mass on Earth, the mass of any object on the moon is <em>precisely identical</em> to its mass on Earth, since mass doesn't depend on where the object is, or on anything else in the environment outside the object.

Its <u>WEIGHT</u>, however, on the moon is only about 16% of its weight on Earth.

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In the parallelogram shown, AE = t + 2, CE = 3t − 14, and DE = 2t + 8.
solniwko [45]
The answer would be 48
7 0
4 years ago
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When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

3 0
3 years ago
Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer
sveticcg [70]

by cosine law we know that

c^2 = a^2 + b^2 - 2 abcos\theta

\theta = 180 - 54.3 - 79.1 = 46.6 degree

now using above equation

93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6

4942.03 = b^2 - 85.34 b

b^2 - 85.34b - 4942.03 = 0

by solving above quadratic equation we have

b = 124.9 m

so it is at distance 124.9 m from deer a

4 0
3 years ago
How are astronomers able to observe and study black holes?
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Info: https://science.nasa.gov/astrophysics/focus-areas/black-holes
 

Hope this Helps! (:

5 0
3 years ago
Can any one please help me on this on kinda lost thank you
Elena-2011 [213]
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