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kenny6666 [7]
3 years ago
9

Is the gravitational force greater on a 1-kg piece of iron or on a 1-kg piece of glass? defend your answer?

Physics
1 answer:
HACTEHA [7]3 years ago
4 0

It is the same, there is no difference.

In fact, the force of gravity on an object is equal to the weight of the object, defined as

W=mg

where m is the mass of the object and g=9.81 m/s^2 is the acceleration of gravity, As we can see, the magnitude of this force depends only on the mass of the object: in this problem, both the objects (the piece of iron and the piece of glass) have same mass (1 kg), therefore they have same weight, and same force of gravity.

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Calculate the kinetic energy, in joules of a 1160-kg automobile moving at 19.0 m/s.
Rudiy27

Answer:

Explanation:

K.E=1/2 mv²

m=1160 kg

v=19.0 m/s

so k.e=1/2*1160*(19.0)²

K.E=1/2*1160*361

K.E=1/2*418760

K.E=209380=2.0*10^5 j

7 0
3 years ago
Which process can transfer energy to a solid metal block
gladu [14]
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4 0
4 years ago
Read 2 more answers
1. Which Law? When you are standing up in a subway train, and the train suddenly stops, your body continues to go forward.
riadik2000 [5.3K]

1. Law 1, since there is no other force acting on your body as you stand there, so you will continue to go forward.

2. Law 3, since the swimmer is using opposite forces to propel herself through the water. She generates a force by pushing the water which helps to push her forward.

3. Law 2, since you are giving the motorcycle more energy as a result of the gas being transformed into the energy that helps to accelerate the motorcycle's speed.

6 0
3 years ago
What is the slowest type of radiation?
tekilochka [14]
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3 0
3 years ago
mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan
Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, v_e, for the planet is v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }

Where;

v_e = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, M_E

r = The radius of the planet = 3 × The radius of Earth, R_E

The escape velocity for Earth, v_e_E, is therefore given as follows;

v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }

\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } =  \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E

v_e = 16 × v_e_E

Given that the escape velocity for Earth, v_e_E ≈ 11,186 m/s, we have;

The escape velocity on the planet = v_e ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0
3 years ago
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