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3 years ago

A 40kg rock falls off a cliff that is 50 meters high. How fast is the speed of the rock when it hits the ground

Physics

1 answer:

Oksi-84 [34.3K]3 years ago

7 0

31.3m/s

Explanation:

Given parameters:

Mass of rock = 40kg

Height of cliff = 50m

Unknown:

Speed of rock when it hits ground = ?

Solution:

We are going to use the appropriate motion equation to solve this problem

The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.

Using;

V² = U² + 2gH

V = unknown velocity

U = initial velocity = O

g = acceleration due to gravity = 9.8m/s²

H = height of fall

since the initial velocity of the bodyg is 0

V² = 2gH

V= √2gH = √2 x 9.8 x 50 = 31.3m/s

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

Similar to another Brainly answer :O

5 0

3 years ago

Conduction circulates heat throughout the atmosphere. True or False Help!!

katrin2010 [14]

Energy is transferred between the ground and the atmosphere via conduction.
Since air is a poor conductor, most energy transfer by conduction occurs right at the earth's surface

7 0

3 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

8 0

3 years ago

Identify the temperature that is equivalent to 95°F. Use this formula to convert the temperature

Stells [14]

The answer is 35 degrees Celsius. Hope I helped :) Please vote brainliest.

7 0

3 years ago

Read 2 more answers