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Ksivusya [100]
3 years ago
9

What is the relation between height and energy?

Physics
1 answer:
Evgesh-ka [11]3 years ago
4 0
Since the gravitational potential energy of an object is directly proportional to its height above the zero position, a doubling of the height will result in a doubling of the gravitational potential energy. A tripling of the height will result in a tripling of the gravitational potential energy.
You might be interested in
if the amplitude of sine curve is 4 and the frequency is 1 with no phase shift, what is the function that descirbes this curve?
Vlada [557]

Answer:

y = 4 Sin (2πt)

Explanation:

Amplitude, A = 4

frequency, f = 1

Wave function is given by

y = A sinωt

where, ω is angular frequency

ω = 2 π f = 2 π x 1 = 2π

So, the desired wave function

y = 4 Sin (2πt)

8 0
4 years ago
Determine the change in electric potential energy of a system of two charged objects when a -2.1-C charged object and a -5.0-C c
elena55 [62]

Answer:

Change in electric potential energy ∆E = 365.72 kJ

Explanation:

Electric potential energy can be defined mathematically as:

E = kq1q2/r ....1

k = coulomb's constant = 9.0×10^9 N m^2/C^2

q1 = charge 1 = -2.1C

q2 = charge 2 = -5.0C

∆r = change in distance between the charges

r1 = 420km = 420000m

r2 = 160km = 160000m

From equation 1

∆E = kq1q2 (1/r2 -1/r1) ......2

Substituting the given values

∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)

∆E = 94.5 × 10^9 (3.87 × 10^-6) J

∆E = 365.72 × 10^3 J

∆E = 365.72 kJ

6 0
3 years ago
The magnetic field on the Sun is created by______.
Ber [7]
The answer is actually c hope this helps
( - - )
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| |
8 0
3 years ago
Find the ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance betw
Snezhnost [94]

Answer:

The ratio of the new force over the original force is 16

Explanation:

Recall the formula for the gravitational force between two masses M1 and M2 separated a distance D:

F_G=G\,\frac{M_1\,\,M_2}{D^2}

So now, if the masses M1 and M2 are quadrupled and the distance stays the same, the new force becomes:

F'_G=G\,\frac{4M_1\,\,4M_2}{D^2}=G\,\frac{16\,\,M_1\,\,M_2}{D^2}=16\,\,G\,\frac{M_1\,\,M_2}{D^2}= 16\,\,F_G

which is 16 times the original force.

So the ratio of the new force over the original force is 16

5 0
3 years ago
At what distance along the central perpendicular axis of a uniformly charged plastic disk of radius 0.390 m is the magnitude of
aliina [53]
<span>Radius, the distance from the centre = 0.390
 Electric field is equal to half of the magnitude. E2 = E / 2
 Given
E1 = E2 E1 = k x Q / r^2
  E2 = (k x Q / r2^2) / 2
  Equating the both we get 2 x r^2 = r2^2
 r2 = square root of (2 x r1^2) = square root of (2) x r = 1.414 x 0.390
  r2 = 1.414 x 0.390 = 0.551 m</span>
3 0
3 years ago
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