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3 years ago

_____ is a area of darkness

Physics

2 answers:

katen-ka-za [31]3 years ago

7 0

Answer:

Shadow

Explanation:

<u>Shadow</u> is a area of darkness.

When sunlight falls on an object ot blocks the light from getting onto the surface . Hence, as light does not falls on that region it is dark. Hence, the shadow is the area of darkness.

OLEGan [10]3 years ago

6 0

Answer:

shadow is a area of darkness

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Mary and Jane are standing at each end of an ice board 1.0 m in length. 2 points

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Answer:

0.053 m ang answer my phsiycal

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3 years ago

Two blocks of clay, one of mass 100 kg and one of mass 3.00 kg, undergo a completely inelastic collision. Before the collision o

adoni [48]

Answer:

a) v = 0.8 m / s , b) $v_{f}$ = 0.777 m / s , c) ΔK = 0.93 J

Explanation:

This exercise can be solved using the concepts of moment, first let's define the system as formed by the two blocks, so that the forces during the crash have been internal and the moment is conserved.

They give us the mass of block 1 (m1 = 100kg, its kinetic energy (K = 32 J), the mass of block 2 (m2 = 3.00 kg) and that it is at rest (v₀₂ = 0)

Before crash

po = m1 vo1 + m2 vo2

po = m1 vo1

After the crash

$p_{f}$ = (m1 + m2) $v_{f}$

a) The initial speed of the block of m1 = 100 kg, let's use the kinetic energy

K = ½ m v²

v = √2K / m

v = √ (2 32/100)

v = 0.8 m / s

b) The final speed,

p₀ = $p_{f}$

m1 v₀1 = (m1 + m2) $v_{f}$

$v_{f}$ = m1 / (m1 + m2) v₀₁

The initial velocity is calculated in the previous part v₀₁ = v = 0.8 m / s

$v_{f}$ = 100 / (3 + 100) 0.8

$v_{f}$ = 0.777 m / s

c) The change in kinetic energy

Initial K₀ = $K_{f}$

K₀ = 32 J

Final $K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.777²

$K_{f}$ = 31.07 J

ΔK = $K_{f}$ - K₀

ΔK = 31.07 - 32

ΔK = -0.93 J

As it is a variation it could be given in absolute value

Part D

For this part s has the same initial kinetic energy K = 32 J, but it is block 2 (m2 = 3.00kg) in which it moves

d) we use kinetic energy

v = √ 2K / m2

v = √ (2 32/3)

v = 4.62 m / s

e) the final speed

v₀₂ = v = 4.62 m/s

p₀ = m2 v₀₂

m2 v₀₂ = (m1 + m2) $v_{f}$

$v_{f}$ = m2 / (m1 + m2) v₀₂

$v_{f}$ = 3 / (100 + 3) 4.62

$v_{f}$ = 0.135 m / s

f) variation of kinetic energy

$K_{f}$ = ½ (m1 + m2) $v_{f}$ ²

$K_{f}$ = ½ (3 + 100) 0.135²

$K_{f}$ = 0.9286 J

ΔK = 0.9286-32

ΔK = 31.06 J

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