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2 years ago

Find the relationship between 'g' and 'G'

2 answers:

6 0

G is the gravitational acceleration while G is the gravitational constant. Both variables are related by the equation g = GM/r^2 where M is mass of an object and r is the distance from the centre of the object.

mel-nik [20]2 years ago

4 0

Lower case and upper case

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During 4 hours one winter afternoon, when the outside temperature was 11° C, a house heated by electricity was kept at 24° C wit

Masteriza [31]

Answer:

Detailed solution is given below:

4 0

3 years ago

Read 2 more answers

If the temperature is held constant during this process and the final pressure is 683 torrtorr , what is the volume of the bulb

Anna [14]

Answer:

Explanation:

Let the volume of the unknown bulb = X L

The volume of the system , after opening valve = (X + 0.72 L )

Use Boyles law gas equation,

P1V1 = P2V2 ( at temperature is constant )

Given:

P1 = 1.2 atm

P2 = 683 torr

Converting mmHg to atm,

1 atm = 760 mmHg(torr)

683 mmHg = 683/760

= 0.8987 atm

1.2X = 0.8987*(X + 0.720)

1.2X = 0.8987X + 0.6471

0.3013X = 0.6471

X = 2.15 L

5 0

3 years ago

A glider with mass m = 0.230 kg sits on a frictionless horizontal air track, connected to a spring with force constant k = 4.50

loris [4]

Answer

given,

mass of glider = 0.23 Kg

spring constant = k = 4.50 N/m

spring stretched to 0.130 m

The springs potential energy =

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 4.5 \times 0.13^2$

U = 0.038 J

at x = 0,the only energy will be kinetic .

$\dfrac{1}{2}mv^2=0.038$

$\dfrac{1}{2}\times 0.23 \times v^2=0.038$

v² = 0.3304

v = 0.575 m/s

displacement of the glider

using conservation of energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$x =v\sqrt{\dfrac{m}{k}}$

$x =3\times \sqrt{\dfrac{0.23}{4.5}}$

x = 0.678 m

8 0

3 years ago

a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

erma4kov [3.2K]

The emerging velocity of the bullet is 71 m/s.

The bullet of mass m moving with a velocity u has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed v when it emerges from the block.

If the block exerts a resistive force F on the bullet and the thickness of the block is x then, the work done by the resistive force is given by,

$W=Fx$

This is equal to the change in the bullet's kinetic energy.

$W=Fx=\frac{1}{2} m(u^2-v^2)......(1)$

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v₁.

Assuming the same resistive forces to act on the bullet,

$F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)$

Divide equation (2) by equation (1) and simplify for v₁.

$\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s$

Thus the speed of the bullet is 71 m/s

3 0

3 years ago

White raven [17]

Answer:

The direction of defliection of the site to the left I think ..

4 0

3 years ago

Find the relationship between 'g' and 'G'​

Find the relationship between 'g' and 'G'