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joja [24]
3 years ago
15

Why centre of mass equal to centre of gravity

Physics
2 answers:
Maksim231197 [3]3 years ago
4 0
Because mass and distance determine gravity, so the more mass you have, the more gravity.
melisa1 [442]3 years ago
4 0
Hopefully this helps you The center of gravity is based on weight, whereas the center of mass is based on mass. So, when the gravitational field across an object is uniform, the two are identical. ... More practically, the COG is the point over which the object can be perfectly balanced; the net torque due to gravityabout that point is zero.
You might be interested in
An object is moving east, and its velocity changes from 65 m/s to 25 m/s in 10 seconds. Which describes the acceleration? negati
hammer [34]

Answer:

we could use the formula, v=u+at,

65=25+a (10), a=4 , since the motion is declerating we have a=-4 m/s2

5 0
3 years ago
Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and
djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

x_f = Final length of the spring .

= \sqrt{1.25^2+1.8^2}  -0.60

= 1.591 m

x_i= The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

=  1.80 m.

m= The mass of the collar.

=3.55 kg

v_c = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c}  = mgy

Or, \frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2}  = 3.55\times 9.81\times 1.80

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)=  8.641 N

Learn more about friction:

brainly.com/question/24338873

#SPJ4

Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

6 0
2 years ago
Momentum is a vector quantity that depends
MissTica
On the change in potential energy
3 0
3 years ago
A skateboarder drops in off the top of one side of the half pipe shown below. She does not push off and starts from rest. She st
solong [7]

Answer:

v

Explanation:

4 0
2 years ago
A child who is swimming toward shore at 0.78 m/s sees shark and picks up his speed
Gnoma [55]

Answer:

0.085m/s²

Explanation:

Use v²=v0²+2a(d)

solve for a

v²-v0²/2d=a

Plug in givens

1.89²-0.78²/2*17.5=a

Plug into calculator

a=0.085m/s²

3 0
3 years ago
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