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Vikentia [17]
3 years ago
12

A 9 kg blob moving at 5 m/s collides and sticks to a second blob with a mass of 5 kg. If the two bobs stick together, what will

the velocity of the two blobs move at?
Physics
1 answer:
vazorg [7]3 years ago
6 0

Answer:

the final velocity of the two blobs after collision is 3.21 m/s.

Explanation:

Given;

mass of first blob, m₁ = 9 kg

mass of the second blob, m₂ = 5 kg

initial velocity of the first blob, u₁ = 5 m/s

initial velocity of the second blob, u₂ = 0

Let the final velocity of the two blobs after collision = v

Apply the principle of conservation linear momentum to determine v;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

9 x 5   +    5 x 0 = v(9 + 5)

45 = 14v

v = 45 / 14

v = 3.21 m/s

Therefore, the final velocity of the two blobs after collision is 3.21 m/s.

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How much time does it take a trucker to deliver his shipment 480 km away travelling at an average speed of 120 km/hr?​
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Answer: time=4 hours

Explanation:

formula:

speed = \frac{distance}{time}

time = \frac{distance}{speed}

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8 0
2 years ago
I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

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period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

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E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

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