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3 years ago

a mango fruit drop down from top of its tree which is 5m high. How long does it take to reach the ground

Physics

2 answers:

Maksim231197 [3]3 years ago

7 0

Answer:

What is the mass tho?

You need the mass to answer this question

nikklg [1K]3 years ago

7 0

Answer:

$distance = ut + \frac{1}{2} g {t}^{2} \\since \: the \: fruit \: is \: from \: rest : \: initial \: velocity \: is \: 0 \: (u = 0) \\ 5 = (0 \times t) + ( \frac{1}{2} \times 9.81 \times {t}^{2} ) \\ 5 = 4.905 {t}^{2} \\ t = \sqrt{1.0194} \\ t = 1 \: second$

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3 years ago

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3 years ago

Amanda spent 2/5 of her time after school doing homework and ¼ of her remaining time riding her bike. If she rode her bike for 4

Doss [256]

Answer:120 min

Explanation:

Given

Amanda spent $\frac{2}{5}$ of her time after school doing Home work

And $\frac{1}{4}$ of her remaining time riding her bike

It is given that she rode her bike for 45 minutes in a week

Let t be the time after school

therefore Amanda spend $\frac{2t}{5}$ in home work and $\frac{3t}{5}$ time is left

From remaining $\frac{3t}{5}$ time she spends $\frac{1}{4}$ time riding her bike

therefore $\frac{3t}{5}\times \frac{1}{4}=45$

thus $t=300 min$

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3 years ago

Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a

Naya [18.7K]

a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

$p_i = m u = (8 kg)(2 m/s)=16 kg m/s$

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

$p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s$

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

$p_f = p_{fB}+p_{fS}$

where $p_{fS}$ is the momentum of the spring. For the conservation of momentum,

$p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s$

b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

$\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s$

c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

$\Delta p=F \Delta t$

Since we know $\Delta t=0.5 s$ , we can find the magnitude of the force:

$F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N$

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.

d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

5 0

3 years ago

a mango fruit drop down from top of its tree which is 5m high. How long does it take to reach the ground​

a mango fruit drop down from top of its tree which is 5m high. How long does it take to reach the ground