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3 years ago

An object is dropped from a 145 m ledge. How fast it is moving just before it hits the ground

Physics

1 answer:

LuckyWell [14K]3 years ago

8 0

D = 1/2 x g × t²
145 = 1/2 × 9.8 × t²
t = 5.4 s

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A horse canters away from its trainer in a straight line, moving 37 m away in 9.1 s. It then turns abruptly and gallops halfway

vesna_86 [32]

Answer:

Average velocity = 1.69 m/s

Average speed = 5.09 m/s

Explanation:

Given that

Horse cover 37 m in 9.1 sec and 18.5 m in 1.8 sec.

As we know that

Average velocity = Displacement / Total time

Average speed = Total distance / Total time

Average velocity:

Average velocity = Displacement / Total time

Total displacement = 37 - 18.5 = 18.5 m

Total time = 9.1 + 1.8 =10.9 s

Average velocity = Displacement / Total time

Average velocity = 18.5 / 10.9

Average velocity = 1.69 m/s

Average speed:

Average speed = Total distance/ Total time

Total distance = 37 + 18.5 = 55.5 m

Total time = 9.1 + 1.8 =10.9 s

Average speed = Total distance/ Total time

Average speed = 55.5 / 10.9

Average speed = 5.09 m/s

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3 years ago

Why should spray cans not be heated?

inna [77]

The can will burst if you do

5 0

4 years ago

Read 2 more answers

Problem 4: A mass m = 1.5 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring

lana [24]

Answer:

y (t) = 0.754 * cos ( 7.96 t - 69.52)

Explanation:

Given:

m = 1.5 kg , k = 95 N / m , v₀ = 6 m / s , d = 0.35 m , t = 0

y (t) = A * cos ( ω * t - φ )

Using the equation that describe the motion

m * v = - k * x ⇒ m * x'' = - k * x

Angular velocity is equal to

ω = √ k / m ⇒ ω = √ 95 N /m / 1.5 kg

ω = 7.96 rad /s

A = v / ω ⇒ A = 6 m /s / 7.96 rad / s

A = 0.754

d = cos * φ ⇒ φ = cos ⁻¹ * 0.35

φ = 69.52

y (t) = A * cos ( ω * t - φ ) ⇒ y (t) = 0.754 * cos ( 7.96 t - 69.52)

4 0

3 years ago

Many people keep a compost pile in their backyards. They use the compost as fertilizer for their flower beds. Which of the follo

olasank [31]

Answer:

I would think B for the answer

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3 years ago

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When you jump upward, your hang time is the time your feet are off the ground. Does hang time depend on the vertical component o

hjlf

Answer:

It only depends on the vertical component

Explanation:

Hello!

The horizontal component will tell you how much you travel in that direction.

You could have a large horizontal velocity, but if the vertical velocity is zero, you will never be out of the ground. Similarly, you could have a zero horizontal velocity, but if you have a non-zero vertical velocity you will be some time off the ground. This time can be calculated by two means, one is using the equation of motion (position as a function of time) and the other using the velocity as a fucntion of time.

For the former you must find the time when the position is zero.

Lets consider the origin of teh coordinate system at your feet

y(t) = vt - (1/2)gt^2

We are looking for a time t' for which y(t')=0

0 = vt' - (1/2)gt'^2

vt' = (1/2)gt'^2

The trivial solution is when t'=0 which is the initial position, however we are looking for t'≠0, therefore we can divide teh last equation by t'

v = (1/2)gt'

Solving for t'

t' = (2v/g)

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4 years ago