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vlabodo [156]
3 years ago
10

A stone fell from the top of a cliff into the ocean. In the air, it had an average speed of 161616 \text{m/s}m/sstart text, m, s

lash, s, end text. In the water, it had an average speed of 333 \text{m/s}m/sstart text, m, slash, s, end text before hitting the seabed. The total distance from the top of the cliff to the seabed is 127127127 meters, and the stone's entire fall took 121212 seconds.
Physics
2 answers:
givi [52]3 years ago
6 0

Answer: It takes the stone 7seconds in air and 5seconds in water

Explanation: complete question: How long did the stone fall in air and how long did it fall in the water?

total distance = sum of velocity * time for both legs of the journey;

let the time taken in air be t

hence time taken in water is 12-t

127 = 16*t + 3(12-t)

127 = 16t + 36 - 3t

127 - 36 = 16t - 3t

91 = 13t

t = 91/13 = 7 seconds 

Hence time taken in air = 7 seconds

time taken in water is 12 - 7 = 5 seconds

 

Vladimir [108]3 years ago
6 0

Answer:

The time it takes to travel in air is 7 seconds

The distance it travels in air is 112 m

The time it takes to travel in water is 5 seconds

The distance it travels in water is 15 m

Explanation:

Let t (s) be the time it takes for the stone to fall in the air. Since the total time is 12 s, then 12 - t is the time it takes to fall in the ocean before hitting the seabed.

The distance it would travel in air in t seconds at the rate of 16 m/s is 16t

The distance it would travel in water in 12 - t seconds at the rate of 3m/s is 3(12-t) = 36 - 3t

The total distance from the cliff to the seabed is 127m, which is 16t + 36 - 3t = 36 + 13t = 127

13t = 127 - 36 = 91

t - 91 / 13 = 7 seconds

The time it takes to travel in air is 7 seconds

The distance it travels in air is 7*16 = 112 m

The time it takes to travel in water is 12 - 7 = 5 seconds

The distance it travels in water is 5*3 = 15 m

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Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, m_1=5\ kg

Initial speed of block A, u_1=5\ m/s

Mass of block B, m_2=8\ kg

Initial speed of block B, u_2=0

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

8 0
3 years ago
An object with a mass of 10 kg is accelerated upward at 2 m/sec2. What force is required?
telo118 [61]

Answer:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

Explanation:

Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.

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Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 31.5°c.
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The rms speed can be calculated using the following rule:
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Substitute with the givens to get the rms speed as follows:
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5. The analytical method of adding vectors expressed in terms of their components may be applied to vectors in three dimensions,
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Answer:

C = 17 i^ - 7 j^ + 16 k^ ,   | C| = 24.37

Explanation:

To work the vactor component method, we add the sum in each axis

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3 0
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