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Natali5045456 [20]
3 years ago
7

What two forces act on falling bodies?

Physics
1 answer:
Alex Ar [27]3 years ago
7 0

The two forces are: Air resistance and gravity.

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When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro
alexandr402 [8]

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

h=ut+\frac{1}{2}at^2

where  

h=displacement

u=initial\ velocity

a=acceleration

t=time

here u=0 because it dropped from a certain height

h=\frac{1}{2}gT^2

T=\sqrt{\frac{2h}{g}}

When height is increases to three times of original height

i.e. h'=3 h

then time period becomes

T'=\sqrt{\frac{2\times 3h}{g}}

T'=\sqrt{3}T

5 0
2 years ago
Which law says that the induced voltage is directly proportional to the magnetic flux?
Vsevolod [243]
law of electromagnetic induction hope this helps 
7 0
3 years ago
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BARSIC [14]

Answer:

b

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

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2 years ago
Which statement best describes frustration?
wlad13 [49]
C. when you can't achieve your goal due to events beyond your control
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2 years ago
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