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damaskus [11]
3 years ago
8

How come that there is a presence of cos there?

Physics
1 answer:
liubo4ka [24]3 years ago
8 0

Answer:

Can i have more context

Explanation:

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Suppose a ball is dropped from shoulder height, falls, makes a perfectly elastic collision with the floor, and rebounds to shoul
Bezzdna [24]

Answer:Same magnitude

Explanation:

When ball is dropped from shoulder height h then velocity at the bottom is given by

v_1=\sqrt{2gh}

if it makes elastic collision then it will acquire the same velocity and riser up to the same height

If m is the mass of ball then impulse imparted is given by

J=m(v_2-v_1)

J=2m\sqrt{2gh}

Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.

7 0
3 years ago
If you saw a waxing gibbous moon what phase would you expect one week later?
Leona [35]
Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.
6 0
3 years ago
Train car A is at rest when it is hit by train car B. The two cars, which have the same mass, stick together and move off after
Natali [406]

Answer:C The final velocity is half of trained car B's initial velocity.

Explanation:

7 0
3 years ago
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Value of g in CGS system
koban [17]

Answer:

in CGS system G is denoted as gram

7 0
3 years ago
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A conducting wire is quadrupled in length and tripled in diameter.
aivan3 [116]

Answer:

Its resistance decreases

Explanation:

The resistance of a wire is directly proportional to the length of the wire and inversely proportional to the cross sectional area of the wire.

Mathematically,

R1 =¶L1/A1... (1)

R1= ¶L1/{Πd²/4}

R1= 4¶L1/Πd²

where;

¶ is the constant of proportionality which is the resistivity of the material

L is the length of the wire

A is the cross sectional area

A1 = Πd²/4

If the length is quadrupled and its diameter tripled

The new length L2 will be 4L1

New area A2 = Π(3d)²/4 = 9Πd²/4

The resistance will become

R2 = ¶(4L1)/{9Πd²/4}

R2 = 4¶L1×4/9Πd²

R2 = 16¶L1/9Πd²... (2)

R2/R1 = 16¶L1/9Πd²÷4¶L1/Πd²

R2/R1 = 16¶L1/9Πd²×Πd²/4¶L1

R2/R1 = 16/9×1/4

R2/R1 = 16/36

R2/R1 = 4/9

R2 = 4/9R1

This shows that the resistance of the wire decreases

7 0
3 years ago
Read 2 more answers
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