All categories

2 years ago

A rock is dropped from a 134-m-high cliff. how long does it take to fall (a) the first 67.0 m and (b) the second 67.0 m?

Physics

1 answer:

____ [38]2 years ago

4 0

A rock is dropped from a 100-m-high cliff. How long do the first 50 meters and the second 50 meters take to fall?

(a) The first 50 meters:

h1 = 0.5 x a x t12

50 m = 0.5% x 9.8% x t12

t1 = 3.19 s

50 meter fall time equals 3.19 seconds.

(a) Counting all 100 meters:

h2 = 0.5 x a x t22

100 m = 0.5% x 9.8% x t22

t2 = 4.52 s

Time to fall 50 meters is equal to t2 - t1 = 1.33 s.

<h3>What is meter?</h3>

The Earth's circumference is around 40000 km, according to the original definition of the metre in 1793, which was one ten-millionth of the distance between the equator and the North Pole on a great circle. The metre was redefined in 1799 using a model metre bar (the actual bar used was changed in 1889). The definition of the metre in terms of a certain number of wavelengths of a specific krypton-86 emission line was made in 1960.

To know more about meter, visit;

brainly.com/question/14475998

#SPJ4

You might be interested in

How can energy be traced as it is converted<br> through different transformations?

Mumz [18]

Answer:

Energy transformations are processes that convert energy from one type (e.g., kinetic, gravitational potential, chemical energy) into another. Any type of energy use must involve some sort of energy transformation.

8 0

4 years ago

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

never [62]

Answer:

1) Vf = 3.36 m/s

2) v = 6.86 m/s

3) s = 92.3 m

4) a = - 0.23 m/s²

Explanation:

We use first equation of motion in this case:

Vf = Vi + at

where,

Vf = Final velocity = ?

Vi = Initial velocity = 0 m/s

a = acceleration = 1.4 m/s²

t = time = 2.4 s

Therefore,

Vf = 0 m/s + (1.4 m/s²)(2.4 s)

We again use first equation of motion but with t= 4.9 s now:

Vf = 0 m/s + (1.4 m/s²)(4.9 s)

Vf = 6.86 m/s

Now, this velocity remained constant for net 11 seconds. Hence, the velocity of hare after 8.9 s is:

First we use second equation of motion to find distance covered in accelerated motion:

s₁ = Vi t + (0.5)at²

s₁ = (0 m/s)(4.9 s) + (0.5)(1.4 m/s²)(4.9 s)²

s₁ = 16.8 m

Now, we calculate the distance covered in uniform motion:

s₂ = vt

s₂ = (6.86 m/s)(11 s)

s₂ = 75.5 m

Now, total distance covered before slowing down is given as:

s = s₁ + s₂ = 16.8 m + 75.5 m

Using third equation of motion for the decelerated motion:

2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance covered = 102 m

Vf = Final Speed = 0 m/s

Vi = Initial Speed = 6.86 m/s

Therefore,

(2)(a)(102 m) = (0 m/s)² - (6.86 m/s)²

a = - (47.06 m²/s²)/(204 m)

3 0

4 years ago

Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m

OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

$E_m = \frac{1}{2} m\omega_1^2A_1^2$