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Georgia [21]
2 years ago
12

NASA sends an unmanned lander to test conditions on the surface of Mars. What is the magnitude of the gravitational force acting

on the lander on the surface of Mars? DATA: Mass of Mars = 6.42·1023 kg Radius of Mars = 3397 km Mass of lander = 138.1 kg
Physics
1 answer:
kozerog [31]2 years ago
4 0

Answer:

The magnitude of the gravitational force acting on the lander on the surface of Mars is 512.46 N.

Explanation:

The Universal law of gravity is define as:

F = G \frac{M.m}{R^{2}}  (1)

Where F is the gravitational force, G is gravitational constant, M is the mass of Mars, m is the mass of the lander and R is the radius of Mars.

Before replacing the values in equation 1 it is necessary to express the radius of mars in terms of meters:

R = 3397 kmx\frac{1000 m}{1 km} ⇒ R = 3397000 m

Finally, equation 1 can be used:

F = (6.67x10^{-11} N.m^{2}/Kg^{2}) \frac{(6.42x10^{23} kg)(138.1 kg)}{(3397000 m)^{2}}

F = 512.46 N

Hence, the magnitude of the gravitational force acting on the lander on the surface of Mars is 512.46 N.

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Five different forces act on an object. Is it possible for the net force on the object to be zero?
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No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
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3 years ago
If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric
poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0
2 years ago
If the magnitude of a charge is half as much as another charge, but the force experienced is the same, then the electric field s
Kazeer [188]

Answer:

the electric field strength of this charge is two times the strength of the other charge

Explanation:

Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;

E1/E2 = Q2/Q1

E2 = E1 x Q1/Q2

      = E x Q/ (Q/2)

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8 0
2 years ago
A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi
Valentin [98]

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

5 0
3 years ago
is dimensionally correct relation necessarily to be a correct physical relation? explain with example.​
Andreas93 [3]

Answer: hope it helps you...❤❤❤❤

Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.

No, not necessarily.

For instance, Newton’s 2nd law is  F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.

But take a look at this (incorrect) equation for the force of gravity:

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It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for  G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.

It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.

A simpler counter example is  1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.

4 0
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