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lawyer [7]
3 years ago
14

A group of engineers have launched a space probe from earth that will land on the surface of Mars. how will the force of gravity

from mars affect the space probe?
A) the force of gravity from Mars will not pull on the space probe until it reaches the planets surface.

B) the force of gravity from Mars will stop pulling on the space probe when it reaches the planet

C) the force of gravity from Mars will pull more strongly on the space probe as it gets closer to Mars.

D) the force of gravity from Mars will pull on the space probe the same amount no matter how far away it is​
Physics
1 answer:
Ierofanga [76]3 years ago
6 0

Answer:D

Explanation:

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3 years ago
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A 40.0 turn coil of wire of radius 3.0 cm is placed between the poles of an electromagnet. The field increases from 0 to 0.75 T
nadezda [96]

Answer:

Induced emf, V=3.76\times 10^{-4}\ V

Explanation:

We have,

Number of turns in the coil, N = 40

Radius of coil, r = 3 cm = 0.03 m

The field increases from 0 to 0.75 T at a constant rate in a time interval of 225 s.

It is required to find the magnitude of the induced emf in the coil if the field is perpendicular to the plane of the coil. The induced emf is given by :

\epsilon=\dfrac{d\phi}{dt}

\phi=NBA\cos\theta, is magnetic flux

\epsilon=NA\dfrac{dB}{dt}\\\\\epsilon=\dfrac{40\times \pi (0.03)^2\times (0.75-0)}{225}\\\\\epsilon=3.76\times 10^{-4}\ V

So, the magnitude of induced emf is 3.76\times 10^{-4}\ V.

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3 years ago
What is numbers 2 and 4? Please help, and many thanks to those who do chose to help.
Elan Coil [88]

Answer:

2. 6 km/h

4. 150 miles/h

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3 years ago
A car of mass m=1000kg is traveling at speed v and brakes. The skid marks are 20m long and the coefficient of kinetic friction i
scoray [572]

Answer:

v = 14 m/s

  = 31.3 mph

The answer would be the same if the mass of the car were 2000 kg

Explanation:

Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.

V^2 = v^2 + 2×a×Δx

Now V, the final velocity is zero as the car stops

0 = v^2 + 2×a×Δx

v^2 = -2×a×Δx

v =√-2×a×Δx    .....*

Now applying Newton's Second Law

Fnet = m×a

-Fk = m×a

-μ×N = m×a

-μ×m×g = m×a (The mass cancels out)

a =  -μ×g

Substituting the value of a back to equation *

v = √-2×(-μ×g)×Δx  

v = √-2×(-0.5×9.8)×20

v = 14 m/s

Therefore the speed the car was travelling with v = 14 m/s

which is equal to 31.3 mph

Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.

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