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a_sh-v [17]
3 years ago
5

The acceleration due to gravity on the surface of Jupiter is about 2.5 times the acceleration due to gravity on Earth’s surface.

What would be the weight of a space probe on the surface of Jupiter?
A.2.5 times lighter than on Earth
B.6.25 times heavier than on Earth
C.2.5 times heavier than on Earth
D.6.25 times lighter than on Earth
Physics
1 answer:
Ann [662]3 years ago
6 0

Answer: The correct answer is option C.

Explanation:

Weight = Mass × Acceleration

Let the mass of the space probe be m

Acceleration due to gravity on the earth = g

Weight of the space probe on earth = W

W=m\times g

Acceleration due to gravity on the Jupiter = g' = 2.5g

Weight of the space probe on earth = W'

W'=mg'=m\times 2.5g

\frac{W'}{W}=\frac{m\times 2.5g}{m\times g}

W'=2.5\times W

The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.

Hence, the correct answer is option C.

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What type of tissue in the heart pumps blood throughout the body?
slavikrds [6]

Answer:

Myocardium. That is the type. (srry i was in a rush hope this helps)

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3 years ago
A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t
Arisa [49]

A) 2.4\cdot 10^{-16}kg

The radius of the oil droplet is half of its diameter:

r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m

Assuming the droplet is spherical, its volume is given by

V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3

The density of the droplet is

\rho=885 kg/m^3

Therefore, the mass of the droplet is equal to the product between volume and density:

m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg

B) 1.5\cdot 10^{-18}C

The potential difference across the electrodes is

V=17.8 V

and the distance between the plates is

d=11 mm=0.011 m

So the electric field between the electrodes is

E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

qE=mg

So, from this equation, we can find the charge of the droplet:

q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

q=-1.5\cdot 10^{-18}C

we can think this charge has made of N excess electrons, so the net charge is given by

q=Ne

where

e=-1.6\cdot 10^{-19}C is the charge of each electron

Re-arranging the equation for N, we find:

N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9

so, a surplus of 9 electrons.

3 0
3 years ago
The volume of a gas is reduced from 4 l to 0.5 l while the temperature is held constant. How does the gas pressure change?
shusha [124]

Answer:

The pressure increases by a factor 8

Explanation:

For a gas held at constant temperature, Boyle's law can be applied. It states that the product of the gas pressure and the gas volume is constant, so we can write:

p_1 V_1 = p_2 V_2

where

p_1 is the initial pressure

p_2 is the final pressure

V_1 is the initial volume

V_2 is the final volume

For the gas in this problem, the volume is reduced from

V_1 = 4 L

to

V_2 = 0.5 L

so we can rewrite the equation as

\frac{p_2}{p_1}=\frac{V_1}{V_2}=\frac{4 L}{0.5 L}=8

this means that the pressure of the gas will increase by a factor 8.

5 0
4 years ago
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