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gayaneshka [121]
3 years ago
6

An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,

with an initial velocity of 3.00j. Find magnitude and direction of the final velocity of the composite.
Physics
1 answer:
Alexus [3.1K]3 years ago
7 0

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       v=\sqrt{2.14^2+0.34^2}=2.17m/s

Direction,  

       \theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

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b.only when the current in the first coil changes.

Explanation:

An induced current flow in the second coil only when there is a change in current in the first cool. A steady current will produce no change in flux (due to magnetic effect of a current) by the first coil, and according to Faraday, induced current is only produced when there is a change in flux linkage.

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3 years ago
A child in a tree house uses a rope attached to a basket to lift a 24 N dog upward through a distance of 4.9 m into the house.
jenyasd209 [6]
<h3>Answer:</h3>

117.6 Joules

<h3>Explanation:</h3>

<u>We are given;</u>

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  • Distance upward is 4.9 m

We are required to calculate the work done

  • Work done is the product of force and distance
  • That is; Work done = Force × distance
  • It is measured in Joules.

In this case;

Force applied is equivalent to the weight of the dog.

Work done = 24 N × 4.9 m

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3 years ago
An 80-cm uniform 10-kg bar is resting on two scales, one at either end. A smaller 4-kg mass (m) is placed at a distance of d = 2
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Answer

given,

length of bar = 80 cm

mass of the bar = 10 kg

smaller mass = 4 kg

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s_1 = 8\ N

s_2 = 14 - s_1

s_2 = 14 - 8

s_2 = 6 N

difference between two scale = 8 - 6

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