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3 years ago

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On Mars gravity is one-third that on Earth. What would be the mass on Mars of a person who has a mass of 90 kilograms (kg) on Ea

rth?

1 answer:

snow_tiger [21]3 years ago

4 0

Answer: The person will still have a mass of 90kg on Mars

Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.

In this case the person will have a Weight of 90*9.8 = 882N on Earth.

{ "g" on Earth is 9.8m/s²}

And a Weight of 90*3.3 = 297N on Mars.

{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}

From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.

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What causes water to move from the liquid part of the hyrdrosphere to the cyrosphere?

bezimeni [28]

Answer:

When the liquid moves through the hydrosphere, the water collects into a cloud. When it falls to the earth, turning into snow and sleet collecting in rivers and lakes.

Explanation:

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3 years ago

Read 2 more answers

What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?

Vikentia [17]

Explanation:

Momentum = mass × Velocity

p = 62×73

p =4526

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3 years ago

Which amusement park ride utilizes periodic motion?

tatyana61 [14]

Any ride that oscillates back and forth or moves only in a complete circle utilizes periodic motion.

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3 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

a)

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

b)

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

c)

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

6 0

4 years ago

Calculate the wavelength (in nm) of the blue light emitted by a mercury lamp with a frequency of 6.88 × 1014 Hz. The speed of li

Nezavi [6.7K]

Answer: 430 nm.

Explanation:

The relation of wavelength and frequency is:

Formula used : $\nu=\frac{c}{\lambda}$

where,

$\nu$ = frequency = $6.88\times 10^{14}Hz$

$\lambda$ = wavelength = ?

c = speed of light = $3.00\times 10^{8}m/s$

Now put all the given values in this formula, we get

$6.88\times 10^{14}=\frac{3.00\times 10^{8}m/s}{\lambda}$

$\lambda=\frac{3.00\times 10^{8}m/s}{6.88\times 10^{14}}=0.43\times 10^{-6}m=430m$ $(1nm=10^{-9}m)$

Thus the wavelength (in nm) of the blue light emitted by a mercury lamp is 430 nm.

6 0

3 years ago