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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.
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A car (m = 1302 kg) traveling along a road begins accelerating with a constant acceleration of 1.50 m/s2 in the direction of mot
antiseptic1488 [7]

First we will use the concepts of motion kinetics for which the final speed is defined as shown below,

v_f^2=v_i^2+2as

Here,

v_f= Final velocity

v_i= Initial velocity

a = Acceleration

s = Distance

Replacing,

(35)^2 = v_i^2+2(1.5)(392)

v_i = 7m/s

Using the conservation of energy for kinetic energy we have,

KE = \frac{1}{2}mv_i^2

KE = \frac{1}{2}(1302)(7)^2

KE = 31900J

Therefore the kinetic energy of the car is 31900J

5 0
3 years ago
In a belly-flop diving contest, the winner is the diver who makes the biggest splash upon hitting the water. the size
ad-work [718]

The second diver have to leap to make a competitive splash by 4.08 m high.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

PE = mgh

where, g = 9.81 m/s²

Given is the diver jumps from a 3.00-m platform. one diver has a mass of 136 kg and simply steps off the platform. another diver has a mass of 100 kg and leaps upward from the platform.

The potential energy of the first diver must be equal to the second diver.

P.E₁ = P.E₂

m₁gh₁ = m₂gh₂

Substitute the vales, we have

136 x 3  = 100 x h₂

h₂ = ₂4.08 m

Thus, the second diver need to leap by 4.08 m high.

Learn more about potential energy.

brainly.com/question/24284560

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7 0
2 years ago
A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f
VashaNatasha [74]

Answer:

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34  

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 =  X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + (\frac{L-y}{2}) ) g sinФ - F2 x (\frac{L}{2}) x gsinФ = 0

plug in the  equations of F1 and F2 into the above equation and after simplification, we get:

(L^{2} - y^{2} ) . w = L^{2} . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

(L^{2} - y^{2} ) . w = L^{2} . 5/9 x w

Hence,

(L^{2} - y^{2} ) = \frac{5L^{2} }{9}

\frac{L^{2} - y^{2}  }{L^{2} } = \frac{5}{9}

Taking L^{2} common and solving for \frac{y}{L}, we will get

\frac{y}{L} = 0.66

Hence, the fraction of the length of the rod above water = \frac{y}{L} = 0.66

and fraction of the length of the rod submerged in water = 1 - \frac{y}{L} = 1 - 0.66 = 0.34

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Economy?
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Answer:

I. don't. get. this. question

C. Demand increases

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When light is incident normally on the interface between two transparent optical media, the intensity of the reflected light is
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i need to ask sum so oycixgixigdtdidit

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