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pickupchik [31]
3 years ago
11

The weight of an object on Earth is determined by

Physics
1 answer:
sukhopar [10]3 years ago
7 0
Answer: D. All of the choices
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Hi!
dangina [55]

Answer:

the moe weight you have in the marble, the higher the speed on the way down

Explanation:

4 0
2 years ago
FAST PLEASE What is the change in internal energy if 90 J of thermal energy is added to a
vladimir1956 [14]
A is right because I took the test
5 0
2 years ago
To what potential should you charge a 2.0 μF capacitor to store 1.0 J of energy?
Bess [88]
E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V

You should charge it to 1000 volts to store 1.0 J of energy.
6 0
3 years ago
Define upthrust simply (don't copy)​
EleoNora [17]

Upthrust is the force that is pushing you up. Kinda to stop you from sinking right into the Earth. Upthrust is simply any force that is causing something to be pushed upwards. Kinda like the opposite of gravity, which is pulling you down =]

Hope this helps ❤️

3 0
2 years ago
A newly discovered planet is found to have a density if 2/3pe and a radius of 2RE, where PE and RED are the density and radius o
Liono4ka [1.6K]
Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"

Solution:
The gravitational acceleration at Earth's surface is given by:
g= \frac{GM}{r^2} 
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius

The Earth's mass can be rewritten also as the product between the Earth's density, d, and its volume (the volume of a sphere of radius r):
M=dV=d ( \frac{4}{3} \pi r^3)=  \frac{4}{3} \pi d r^3 

Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
g' =  \frac{GM'}{r'^2} (1)
so we need to find M' and r'.

The problem says the radius of the new planet is twice the Earth's radius: 
r'=2r (2)
and that its density is 2/3 of Earth's density:
d'= \frac{2}{3} d
so the mass M' of the new planet is, with respect to the Earth's mass:
M' = d'V' = \frac{4}{3} \pi d' (r')^3 =  \frac{4}{3} \pi ( \frac{2}{3}d) (2r)^3 = ( \frac{4}{3} \pi d r^3 )( \frac{16}{3}) =  \frac{16}{3} M (3)

And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
g'= \frac{G( \frac{16}{3}M) }{(2r)^2}=  \frac{GM}{r^2}  \frac{4}{3} =  \frac{4}{3}g
And since g=9.81 m/s^2, we find
g'=  \frac{4}{3}(9.81 m/s^2)=13.1 m/s^2
8 0
3 years ago
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