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Art [367]
3 years ago
7

which object has the most gravitational potential energy? A. an 8 kg book at a height of 3m B. an 5 kg book at a height of 3 m C

. A 5 kg book at a height of 2 m D. a book at a height of 2 m
Physics
2 answers:
astra-53 [7]3 years ago
6 0
I think A is the correct answer because its high is more higher compared to the others, and the mass really does not matter, to know the gravitational potential energy, we need to know how high the object is located because gravity does not show any favor to an object that has more mass or an object that doesnt
amid [387]3 years ago
5 0

Answer: an 8 kg book at a height of 3m

Explanation:

Potential energy is the energy possessed by an object by virtue of its position.

P.E=m\times g\times h

m= mass of object = 200 kg

g = acceleration due to gravity = 10ms^{-2}

h = height of an object  

1. an 8 kg book at a height of 3m

P.E=8kg\times 10ms^{-2}\times 3m=240Joules

2. an 5 kg book at a height of 3 m

P.E=5kg\times 10ms^{-2}\times 3m=150Joules

3. an 5 kg book at a height of 2 m

P.E=5kg\times 10ms^{-2}\times 2m=100Joules

4. a book at a height of 2 m

P.E=0kg\times 10ms^{-2}\times 3m=0Joules

Thus a 8 kg book at a height of 3 m has most potential energy.

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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

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A 3.4 nC charged particle has a velocity of 4.7 m/s and is moving in the +x-direction. If this charge is in a magnetic field det
zloy xaker [14]

Answer:

Magnetic force, F=1.22\times 10^{-7}\ N

Explanation:

It is given that,

Charge, q=3.4\ nC=3.4\times 10^{-9}\ C

Velocity, v = 4.7 m/s

Magnetic field, B=-1.4i+7.52j

|B|=\sqrt{(-1.4)^2+(7.52)^2}=7.64\ T

Magnetic force is given by :

F=q\times v\times B

F=3.4\times 10^{-9}\ C\times 4.7\ m/s\times 7.64\ T

F=1.22\times 10^{-7}\ N

So, the magnetic force on this particle is 1.22\times 10^{-7}\ N. Hence, this is the required solution.

5 0
3 years ago
What is the maximum height achieved if a 0.600 kg mass is thrown straight upward with an initial speed of 75.0 m·sâ1? ignore th
loris [4]

We can solve the problem by using the law of conservation of energy:

- at the beginning, all mechanical energy of the object is just kinetic energy: K=\frac{1}{2}mv^2, where m is the mass and v is the velocity

- at the point of maximum height, all mechanical energy of the object is just gravitational potential energy: U=mgh, where h is the maximum height

Therefore, the conservation of energy becomes:

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

Re-arranging, we find the maximum height:

h=\frac{v^2}{2g} = \frac{(75.0 m/s)^2}{2(9.8 m/s^2)}=287.0 m


6 0
3 years ago
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