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zepelin [54]
3 years ago
6

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou

nd PLEASE HELP
Physics
1 answer:
IceJOKER [234]3 years ago
7 0

The potential energy is 4.10 J

Explanation:

The potential energy of a body is given by

U=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the height of the object relative to the ground

For the lemming in this problem we have

m = 0.078 kg is its mass

h = 5.36 m is the height above the ground

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

U=(0.078)(9.8)(5.36)=4.10 J

Learn more about potential energy here:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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Calculate the electric field strength at a point at which a test charge of 0.30 coulombs experiences a force of 5.0 newtons.
SVEN [57.7K]

The strength of electric field E is 17 N / C.

<u />

<u>Explanation:</u>

Electric field strength is defined as the force per unit charge acting at a point in the given field. The equation for the strength of the electric field is given by

                     E = F / q

where E represents the electric field strength,

           F represents the force in newton,

           q represents the charge in coulomb.

Given the charge q = 0.30 coulombs

                   force F = 5.0 N

Electric field strength E = force / charge

                                        = 5.0 / 0.30

                                    E  = 16.66 = 17 N / C.

5 0
3 years ago
What mistake did Dominic make?
Tatiana [17]

Answer:

Im not sure who you are talking about, but if you are talking about Dominic Raab, well, he failed to make a crucial phone call to seek urgent help airlifting translators out of Afghanistan.          if you are not talking about this Dominic, then please add more information to your question.

7 0
2 years ago
The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete
sweet [91]

Answer:

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

F_n = 1.72 \times 10^4

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

v = \sqrt{5 R g}

also by energy conservation this is gained by initial potential energy

mgh = \frac{1}{2}mv^2

v = \sqrt{2gh}

so we will have

\sqrt{2gh} = \sqrt{5Rg}

now we have

h = \frac{5R}{2}

here we have

R = 7.5 m

so we have

h = \frac{5(7.5)}{2}

h = 18.75 m

Now when it will reach at point B then its normal force is just equal to ZERO

N_B = 0

now when it reach point C then the speed will be

mgh - mg(2R_c) = \frac{1}{2]mv_c^2

v_c^2 = 2g(h - 2R_c)

v_c = 13.1 m/s

now normal force at point C is given as

F_n = \frac{mv_c^2}{R_c} - mg

F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)

F_n = 1.72 \times 10^4

7 0
4 years ago
Which of the following are results of the force of gravity?
docker41 [41]
Hello,

Here is your answer:

The proper answer for this question is option B "When released,a book falls to the ground". That's because of gravity the book will hit the ground!

Your answer is B.

If you need anymore help feel free to ask me!

Hope this helps.
7 0
4 years ago
Read 2 more answers
A very thin circular disk of radius R = 20.00 cm has charge Q = 30.00 mC uniformly distributed on its surface. The disk rotates
lutik1710 [3]

Answer:

B= 7.5*10^{-15}T

Explanation:

The magnetic field strenght on the z-axis at a distance d from the center is,

B= \frac{\mu_0 Q\omega R^2}{8\pi d^3}

Our values are:

R=20cm\\Q=30mc\\w=5rad/s\\d=2*10^3cm=20m

Replacing,

B= \frac{(4\pi*10^{}-7)(30*10^{-3})(5)(0.2)^2}{8\pi(20)}

B= 7.5*10^{-15}T

3 0
3 years ago
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