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emmainna [20.7K]
3 years ago
14

If you take rocks from the Earth to the moon (5 points)

Physics
2 answers:
Lemur [1.5K]3 years ago
8 0
the mass would remain the same and the weight would change. mass defines the actual amount of an object, while weight is the force gravity has on that object. gravity is different on the moon and the earth therefore the weight will change
katrin2010 [14]3 years ago
4 0

Answer:

the mass would remain the same and the weight would change

Explanation:

Mass can be thought of as the amount of matter an object has, this quantity is a constant, it will not change even if the rocks are carried to the moon.

What will change is the weight, since it is defined as:

w = mg

where w is the weight, m is the mass and g is the acceleration due to gravity.

We said the mass stays the same, but the acceleration g on the moon will be different (smaller) than the acceleration g on the earth.

So because the weight depends on the acceleration of gravity, if the rocks are taken to the moon where that acceleration is different, the rocks will have a different weight.

So the answer is: the mass would remain the same and the weight would change

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. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of
Rina8888 [55]

The force required is 319 N

Explanation:

The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:

F_f = \mu W

where

\mu is the coefficient of friction

W is the weight of the object

Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

\mu = 0.55 (coefficient of friction)

W=580 N (weight of the leaves)

Substituting, we find:

F=(0.55)(580)=319 N

Learn more about force of friction:

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7 0
4 years ago
Why do the waves have different speeds in different layers of Earth's surface?
Genrish500 [490]

Answer:

Material's density

Explanation:

Seismic waves travel at different rates of speed based on a material's density. Hopefully, you understand that the Earth has three main layers: the crust, mantle, and core. Earthquake waves move faster through solids.

6 0
2 years ago
For the airfoil and conditions in Problem 2.2, calculate the lift-to-drag ratio. Comment on its magnitude.
raketka [301]

Answer:

L/D= 112

Explanation:

Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.

Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.

Lift increases proportionally to the square of the speed.

The solutions to the question is the file attached to this explanation.

Lift,L= qC(l). S---------------------------(1).

and,

Drag,D = qC(d).S ----------------------(2).

Hence, Lift to drag ratio,L/D= C(l)/C(d).

Therefore, we have to compute various angle of attack.(check attached file)...

Then, (L/D) will then be equal to 112.

8 0
4 years ago
According to ohm's law if you don't change the value of the resistor & you double the voltage in a circuit the amount of cur
Nana76 [90]

They double!! Hope im not too late!!

3 0
3 years ago
(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o
IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
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