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3 years ago

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It turns out that Mercury and Mars have the same gravity as one another – that is, you would weigh the same on the surface of Me

rcury as you would on the surface of Mars, despite the fact that Mercury is smaller. Why is this the case?

1 answer:

Tamiku [17]3 years ago

4 0

Answer:

Explanation:

The value of acceleration due to gravity of mars is same as that the value of acceleration due to gravity of Mercury.

The value of acceleration due to gravity of a planet depends on its mass and the radius.

The formula for the acceleration due to gravity is given by

$g=\frac{GM}{R^{2}}$

Where, M be the mass of the planet, R be the radius of the planet.

As according to the question, the value of acceleration due to gravity for Mercury is same as that of Mars.

$\frac{GM_{mercury}}{R_{mercury}^{2}}=\frac{GM_{mars}}{R_{mars}^{2}}$

$\frac{M_{mercury}}{R_{mercury}^{2}}=\frac{M_{mars}}{R_{mars}^{2}}$

As teh radius of Mercury is small, and acceleration due to gravity is same, it is possible because the mass of Mercury is more than the mass of Mars.

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3 years ago

Read 2 more answers

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

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3 0

3 years ago

A 13.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay s

Alisiya [41]

Answer:

$v_{ic}=92.53 m/s$

Explanation:

We need to apply conservation of momentum and energy to solve this problem.

<u>Conservation of momentum</u>

$p_{i}=p_{f}$

$m_{c}v_{ic}=(m_{c}+m_{w})V$ (1)

m(c) is the mass of stick clay
m(w) is the mass of the wooden block
v(ic) is the initial velocity of clay
V is the final velocity of the system clay plus wood.

<u>Conservation of total energy</u>

The change in kinetic energy is equal to the change in internal energy, in our case it would be the energy loss due to the friction force. Let's recall the definition of work, it is the dot product between force and displacement, Therefore:

$\Delta E=W$

$\frac{1}{2}(m_{c}+m_{w})V^{2}=F_{friction}*d$

$\frac{1}{2}(m_{c}+m_{w})V^{2}=\mu (m_{c}+m_{w})gd$

We can find V from this equation:

$V=\sqrt{2\mu gd}=\sqrt{2*0.65*9.81*7.5}=9.78 m/s$

Now, let's put V into the equation (1) and find v(ic)

$v_{ic}=\frac{(m_{c}+m_{w})V}{m_{c}}=\frac{123*9.78}{13}=92.53 m/s$

I hope it helps you!

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8 0

3 years ago