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Leni [432]
3 years ago
15

Amit buys few

Physics
1 answer:
Deffense [45]3 years ago
6 0

Answer:

No, the apparent weight will be different.

Explanation:

Although the mass is still the same, the apparent weight will be different.  This is because of the centripetal acceleration caused by the Earth's rotation.  At the poles, there is no centripetal acceleration.  The closer you get to the equator, the larger the centripetal acceleration becomes, and the lower the apparent weight becomes.

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How and why does air pressure change with altitude in the atmosphere?
wlad13 [49]
Pressure with Height: pressure decreases with incrementing altitude. The pressure at any caliber in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a homogeneous surface at lower calibers.
5 0
3 years ago
A proton is held at rest in a uniform electric field. When it is released, the proton will lose?
nydimaria [60]

A proton is held at rest in a uniform electric field. When it is released, the proton will lose its kinetic energy.

Kinetic energy

The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. The body exerts the same amount of effort when slowing down from its current pace to a condition of rest. Formally, kinetic energy is any term that includes a derivative with respect to time in the Lagrangian of a system.

To learn more about kinetic energy refer here:

brainly.com/question/11301578

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5 0
2 years ago
The force of gravitation between two spherical bodies is Gm1 m2 /r2, where r is separation between their dash
AnnZ [28]

Explanation:

F = Gm1m2/r^2

kya nikalna hai bhai isme

6 0
3 years ago
Read 2 more answers
Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?
Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

A_y=+15 m

\theta=32^{\circ}

We have to find the x-component of A and magnitude of A.

According to question

A_y=\mid A\mid cos\theta

Substitute the values then we get

15=\mid A\mid cos32

\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}

\mid A\mid=17.7m

tan\theta=\frac{perpendicular\;side}{Base}

tan32=\frac{A_x}{A_y}=\frac{A_x}{15}

0.62\times 15=A_x

A_x=9.3

The value of x-component of A is negative because the vector A lie  in second quadrant.

Hence, the x- component of A=-9.3 m

6 0
3 years ago
Read 2 more answers
A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th
gregori [183]

Answer: The center of gravity is 1.1338 m away from the left side of the barbell

Explanation:

Length of the barbell = 1.90 m

The distance center of gravity from left = x

Mass on the left side = 25 kg

The distance center of gravity from right = 1.90 - x

Mass on the right side = 37 kg

At the balance point: m_1x_1=m_2x_2

25 kg\times x=37 kg\times (1.90-x)

x=1.1338 m

The center of gravity is 1.1338 m away from the left side of the barbell

7 0
4 years ago
Read 2 more answers
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