Ask question

Ask question

All categories

3 years ago

13

A ball is thrown horizontally at a speed of 24 m/s from the top of a cliff. if the ball hits the ground 4.0 s later, approximate

ly how high is the cliff?

1 answer:

marishachu [46]3 years ago

6 0

144 m 24 divided by 4 = 6 24 x 6 = 144

You might be interested in

On a Vernier Caliper, how do you know which mark to use on the very top scale?

madreJ [45]

<u>Answer</u>

To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use<em>. </em>

<u>Explanation</u>

A vernier caliper is an instrument that is used to measure the diameter of small circular objects such as diameter of a wires, thickness of an iron sheet.

The objects to be measured is place between the jaws of the calipers.

The vernier scale has two scales, the vernier scale and the main scale which is the very top scale.<em> To know where it starts we look where the zero mark of the vernier scale starts. The make just before reaching where the zero mark is marks the value to use. </em>

4 0

3 years ago

A child rides her bike at a rate of 12.0 km/hr down the street. A squirrel suddenly runs in front of her so she applies the brak

Sedbober [7]

By v = u - at
<span>=>8 = 12 - a x 0.25 </span>
<span>=>a = 4/0.25 km/hr/sec </span>
<span>=>a = 16km/hr/sec

I hope this helped!</span>

3 0

3 years ago

Read 2 more answers

Which of the five themes of geography would a person most likely use to study the importation of Chinese goods into the United S

Morgarella [4.7K]

B. Movement would be the correct answer.

7 0

2 years ago

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago

GIVING BRAINLIEST PLEASE HELP ME!!!

kumpel [21]

Answer:

I'm Pretty sure the answer your looking for is C

4 0

3 years ago