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2 years ago

6

A rock is thrown downward from the top of a cliff If the rock hits the ground after 2.0 s, with a velocity of 12 m/s what is the

height of the cliff?

1 answer:

expeople1 [14]2 years ago

8 0

Explanation:

v = u + a t

12 = 0 + 2a

a= 12/2

a=6 m/s2

v^2 - u ^2 = 2 as

12 ^2 - 0 ^2 = 2 * 6 * s

144 = 12 s

s = 144/12

s = 12 m

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3 years ago

Durning which type of process does pressure remain consistent

Simora [160]

Answer:

$\fbox {D. Isobaric}$

Explanation:

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2 years ago

Read 2 more answers

A 0.450 kg soccer ball has a kinetic energy of 119 J.

Anastaziya [24]

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2 = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

Square root both sides

Sq rt v² = sq rt 528.89

V = 22.998m/s

V is approximately = 23m/s

I hope this was helpful, please rate as brainliest

8 0

2 years ago

What two factors determine how bright a star appears to be in the sky?

Delicious77 [7]

Answer:

1. Luminosity

2.Apparent brightness

Explanation:

There are two factors on which brightness of star appear to be in the sky

The two factors are

1. Luminosity

2.Apparent brightness

1.Luminosity :It is defined as the total energy emitted by the object in a given time.Luminosity vary with the distance of observer from the star.Luminosity is a intrinsic property which depends on the fundamental chemical composition and structure of the material.Luminosity is depends on the size of star.Lager the star luminosity will be more.

2.Apparent brightness: It is defined as how bright a star appears from an observer on the earth and the amount of starlight reaching the earth.if the distance is large then the brightness decreases.When the distance of star from us small then the brightness of star increases.Distance is inversely proportional to brightness of the star.

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3 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

$\omega=-\alpha t$

$\alpha=-\dfrac{\omega}{t}$

$\alpha=-\dfrac{50.0}{20.0}$

$\alpha=-2.5\ rad/s^2$

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

$\theta=\omega_{0}t+\dfrac{1}{2}\alpha t$

Put the value into the formula

$\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2$

$\theta=500\ rad$

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

$\vec{\tau}=\vec{r}\times\vec{f}$

$\tau=r\times f\sin\theta$

Put the value into the formula

$\tau=2.5\times4\times 9.8\sin60$

$\tau=84.87\ N-m$

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

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3 years ago