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DerKrebs [107]
3 years ago
5

On a distant planet, the GPE of a 65 kg astronaut is 11,115 j when they are on a 46 m tall cliff. What is the acceleration due t

o gravity on this planet
Physics
1 answer:
Ainat [17]3 years ago
3 0

Answer:Learn what gravitational potential energy means and how to calculate it. ... a pulley and rope, so the force due to lifting the box and the force due to gravity, ... would be used by an elevator lifting a 75 kg person through a height of 50 m if the ... When you are close to a planet you are effectively bound to the planet by gravity ..

Explanation:

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What is the SI unit for momentum?
Whitepunk [10]
Kg . Meter per second (Kg.m/s)
8 0
3 years ago
Read 2 more answers
Un automóvil acelera de 0 a 140 km/h en 9.8 segundos, determina su aceleración
guajiro [1.7K]

Answer:

The acceleration is 14.28 km/h^2

Explanation:

Step one:

Given data

initial speed u= 0 km/h

final speed v= 140km/h

time t= 9.8 seconds

Required

The acceleration of the car

Step two:

From a= v-u/t

substitute

a= 140-0/9.8

a=140/9.8

a=14.28 km/h^2

6 0
3 years ago
A force of 120.0 N is applied to a 3.00 kg block. What is the average acceleration of the block?
kirza4 [7]

Explanation:

F = ma

120.0 N = (3.00 kg) a

a = 40.0 m/s²

6 0
3 years ago
Two concentric conducting spherical shells produce a radially outward electric field of magnitude49,000 N/C at a point 4.10 m fr
Korolek [52]

To solve this problem it is necessary to apply the concepts related to the electric field according to the definition of Coulomb's law.

 The electric field is defined mathematically as a vector field that associates to each point in space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point.

Mathematically this can be described as:

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

Where,

\epsilon_0 = permittivity of free space

r = Distance

q = Charge

E = Electric Field

Our values are given as,

E= 49.000N/C

r= 4.1m

\epsilon_0=8.854*10{-12}C^2N^{-1}m^{-2}

Replacing we have,

E = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}

49000 = \frac{1}{4\pi (8.854*10{-12})}\frac{q}{(4.1)^2}

q= 9.16*10^{-5} C

q= 91.6\mu C

Therefore the amoun of charge on the outer surface of the larger shell is 91.6 \mu C

6 0
3 years ago
As sources of electrical power, windmills now account for only about 2,500 megawatts nationwide, but production is almost expect
Zanzabum

Answer: C

Explanation:

Production is expected that it will almost double by the end of the year is the answer because production in this context relates to a numerical quantity which is a function of time.

Therefore the probability there is on whether production will almost double to provide enough electricity and not if production will occur.

6 0
3 years ago
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