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mash [69]
3 years ago
10

Calculate potential energy

Physics
1 answer:
IrinaVladis [17]3 years ago
4 0
The potential energy of an object is defined by the equation: PE = mgh, where m = the mass of the object, g = the gravitational acceleration and h = the object's height above the ground.
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If a ball has the weight of 3.76, how much work would it take to lift it 2 meters above the ground? (sorry if this question does
Aloiza [94]
We know, work done = Weight * Displacement 
w = 3.76 * 2
w = 7.52 J

In short, Your Answer would be: 7.52 Joules

Hope this helps!
7 0
2 years ago
what is the magnitude of the gravitational force acting on the earth due to the sun? express your answer in newtons.
Alborosie

The gravitational force the sun experiences from the earth is 3.48×10²²N, which is exactly the same as the force the sun experiences from the earth.

  • Gravity is a force that develops as a result of the attraction between mass-containing objects. The mass of the object has a direct relationship to the strength of this attraction. r equals the separation of two objects.

F = G (M₁M₂)/r²

Where, F  the gravitational force

G=6.67×10⁻¹¹Nm²kg⁻² gravitational constant

M₁=5.98×10²⁴kg  mass of earth

M₂= 1.99×10³⁰ kg the mass of the sun

r =15×10¹⁰ m is the distance between sun and earth

Putting all the values in above equation,

F = 6.67×10⁻¹¹Nm²kg⁻²(5.98×10²⁴kg 1.99×10³⁰ kg)/15×10¹⁰ m

On solving the above equation we get,

F = 3.48×10²²N

To know more about gravitational force

brainly.com/question/12830265

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5 0
1 year ago
The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. If HST has a tangential speed of 7,750 m/s, how long is HS
deff fn [24]

Answer: 5,640 s (94 minutes)

Explanation:

the tangential speed of the HST is given by

v=\frac{2\pi r}{T} (1)

where

2\pi r is the length of the orbit

r is the radius of the orbit

T is the orbital period

In our problem, we know the tangential speed: v=7,750 m/s. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

r=6.38\cdot 10^6 m+569,000 m=6.95\cdot 10^6 m

So, we can re-arrange equation (1) to find the orbital period:

T=\frac{2 \pi r}{v}=\frac{2 \pi (6.95\cdot 10^6 m/s)}{7,750 m/s}=5,640 s

Dividing by 60, we get that this time corresponds to 94 minutes.

6 0
3 years ago
Read 2 more answers
Show that the energy of a magnetic dipole m in a magnetic field B is U--m B
Juli2301 [7.4K]

Answer:

showm

Explanation:

Consider a dipole having magnetic moment 'm' is placed in magnetic field \vec{B} then the torque exerted by the field on the dipole is

\tau = m\times B

\tau=mBsin\alpha

Now to rotate the dipole in the field to its final position the work required to be done is

U=\int \tau d\alpha

U=\int mBsin\alpha d\alpha

U= -mBcos\alpha

U=-\vec{m\times \vec{B}}

Minimum energy mB is for the case when m is anti parallel to B.

Minimum energy -mB is for the case when m is parallel to B.

5 0
3 years ago
The low-pressure area near Earth’s equator is filled by cool air moving in from ________. Btw this is science
bonufazy [111]

Answer:

the north and south pole

Explanation:

this should be the correct answer

5 0
2 years ago
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