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N76 [4]
3 years ago
7

2. What is the standard value of acceleration due to gravity or "g"?

Physics
2 answers:
lawyer [7]3 years ago
8 0

Answer:

9.80665 m/s^2

Explanation:

This is the value on Earth.

svp [43]3 years ago
6 0

Answer:

The standard acceleration due to gravity (or standard acceleration of free fall), sometimes abbreviated as standard gravity, usually denoted by ɡ0 or ɡn, is the nominal gravitational acceleration of an object in a vacuum near the surface of the Earth. It is defined by standard as 9.80665 m/s2 (about 32.17405 ft/s2).

Explanation:

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An ice skater weighs 500 [N]. He is coasting to the right at a constant velocity of 2 [m/s]. Assume
luda_lava [24]

Answer:

The net force on the skater is zero. (F_{net} = 0\,N)

Explanation:

According to Newton's First Law, an object is at equilibrium when either it is at rest or moves at constant velocity, which means a net force of zero. Based on the given statement, there are no external forces acting on skate and, therefore, the net force on the skater is zero. (F_{net} = 0\,N)

4 0
3 years ago
Why is a rotting apple a reaction in which energy is neither absorbed nor released?
o-na [289]

Answer:

Hey

It would have to be C because no net energy is lost.

7 0
3 years ago
1. 3 main components of a circuit
siniylev [52]
Voltage, resistance and current are the three components that must be present for a circuit to exist. A circuit will not be able to function without these three components. Voltage is the main electrical source that is present in a circuit. :)
7 0
3 years ago
Read 2 more answers
The force of gravity on a 60 kg woman is 588 N. The woman also exerts a gravitational force on
bulgar [2K]

Answer:

the same    588 N

Explanation:

3 0
2 years ago
My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin
Tresset [83]

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

U₀ = K₁    

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

8 0
3 years ago
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