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Darya [45]
3 years ago
15

What is the acceleration due to gravity on Mars?

Physics
1 answer:
Leona [35]3 years ago
8 0

The acceleration of gravity on Mars is about 3.7 m/s². (b)

So anything located on the surface of Mars weighs about 38% of what it would weigh if it were on Earth.

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From left to right across a period in the periodic table, elements become less ___________ and more ______________ in their prop
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A 3.3 kg ball sits on the ground and is kicked with a FAPP of 36N
jeyben [28]

a) 32.3 N

The force of gravity (also called weight) on an object is given by

W = mg

where

m is the mass of the object

g is the acceleration of gravity

For the ball in the problem,

m = 3.3 kg

g = 9.8 m/s^2

Substituting, we find the force of gravity on the ball:

W=(3.3)(9.8)=32.3 N

b) 48.3 N

The force applied

F_{app} = 36 N

The ball is kicked with this force, so we can assume that the kick is horizontal.

This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:

F=\sqrt{W^2+F_{app}^2}

And substituting

W = 32.3 N

Fapp = 36 N

We find

F=\sqrt{32.3^2+36^2}=48.3 N

c) 14.6 m/s^2

The ball's acceleration can be found by using Newton's second law, which states that

F = ma

where

F is the net force on an object

m is its mass

a is its acceleration

For the ball in this problem,

m = 3.3 kg

F = 48.3 N

Solving the equation for a, we find

a=\frac{F}{m}=\frac{48.3}{3.3}=14.6 m/s^2

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3 years ago
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A parallel-plate capacitor has a plate area of 0.2 m^2 and a plate separation of 0.1 mm. If the charge on each plate has a magni
barxatty [35]

Answer:

The potential difference across the plates is 226 V.

Explanation:

Given;

area of the capacitor plate, A = 0.2 m²

separation, d = 0.1 mm = 0.1 x 10⁻³ m

charge on each plate, Q = 4 x 10⁻⁶ C

Charge on the capacitor is given by;

Q = CV

Where;

C is the capacitance of the capacitor, given as;

C = ε₀A / d

Then, the potential difference across the plates is given by;

V = \frac{Q}{C} = \frac{Qd}{\epsilon_o A} = \frac{(4*10^{-6})(0.1*10^{-3})}{(8.85*10^{-12})(0.2)}\\\\V = 226 \ V

Therefore, the potential difference across the plates is 226 V.

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3 years ago
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