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A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s . What is the height of the cliff? (Use 32 ft/s 2 for

bulgar [2K]

To solve this problem we will apply the linear motion kinematic equations, which describe the change in velocity, depending on the acceleration and the distance traveled, that is,

$v_f^2 = v_i^2 +2ah$

Where,

$v_f$ = Final Velocity

$v_i$ = Initial Velocity

a = Acceleration

h = height

Our values are given as,

$v_f = 88 ft/s\\v_i = 0 ft /s\\a = 32 ft/s^2\\$

Replacing we have,

$vf^2 = vi^2 + 2*a*h$

$88^2 = 0 + 2*32*h$

$h= 121 ft$

Therefore the height of the cliff is 121ft

5 0

3 years ago

Two fans are watching a baseball game from different positions. One fan is located directly behind home plate, 18.3 m from the b

EleoNora [17]

Answer:

0.317 s

Explanation:

distance of fan A = 18.3 m

distance of fan B = 127 m

speed of sound (s) = 343 m/s

What is the time difference between hearing the sound at the two locations?

time (T) = distance / speed

time for sound to reach fan A = 18.3 / 343 = 0.053 s

time it takes for sound to reach fan B = 127 / 343 = 0.370 s
time difference = 0.370 - 0.053 = 0.317 s

5 0

3 years ago

The objects listed are placed at the top of a ramp and roll down to the bottom without slipping. Assuming that there is no air r

Thepotemich [5.8K]

Explanation:

For each object, the initial potential energy is converted to rotational energy and translational energy:

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For the marble (a solid sphere), I = ⅖ mr².

For the basketball (a hollow sphere), I = ⅔ mr².

For the manhole cover (a solid cylinder), I = ½ mr².

For the wedding ring (a hollow cylinder), I = mr².

If we say k is the coefficient in each case:

mgh = ½ (kmr²) ω² + ½ mv²

For rolling without slipping, ωr = v:

mgh = ½ kmv² + ½ mv²

gh = ½ kv² + ½ v²

2gh = (k + 1) v²

v² = 2gh / (k + 1)

The smaller the value of k, the higher the velocity. Therefore:

marble > manhole cover > basketball > wedding ring

7 0

4 years ago

A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?

Answer:

The value is $T_R = 11.8 \ days$

Explanation:

From the question we are told that

The semi - major axis of the rocky debris $a_R = 45.0\ AU$

The semi - major axis of Planet D is $a_D = 60 \ AU$

The orbital period of planet D is $T_D = 18.164 \ days$

Generally from Kepler third law

$T \ \ \alpha \ \ a^{\frac{3}{2} }$

Here T is the orbital period while a is the semi major axis

So

$\frac{T_D}{T_R} = \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}$

=> $T_R = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }$

=> $T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }$

=> $T_R = 11.8 \ days$

7 0

3 years ago

On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f

aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

gm = 1/6 ge

gm = 1/6 9.8 m/s² = 1.63 m/s²

We calculate the range

R = Vo² sin 2θ / g

R = 25² sin (2 30) / 1.63

R= 332 m

We will calculate the time of flight,

Y = Voy t – ½ g t2

Voy = Vo sin θ

When the ball reaches the end point has the same initial height Y=0

0 = Vo sin  t – ½ g t2

0 = 25 sin (30) t – ½ 1.63 t2

0= 12.5 t – 0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

5 0

3 years ago