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Jlenok [28]
3 years ago
9

a satellite is traveling in a orbit around the earth. what change would cause it to experience more gravitational force from the

earth? A. orbiting closer to the earths surface. B. removing some of its mass. C making it orbit faster. D. making it's surface reflect the sun's rays.
Physics
1 answer:
never [62]3 years ago
7 0
The correct answer is:
<span>A. orbiting closer to the earths surface. 

In fact, the gravitational force exerted by the Earth on the satellite is
</span>F=G \frac{Mm}{r^2}
<span>where
G is the gravitational constant
M is the Earth mass
m is the satellite mass
r is the distance of the satellite from the Earth's surface

We can see that, if the satellite orbits closer to the Earth's surface, its distance r from the centre of the planet decreases. But when r decreases, F (the gravitational force) increases, so A is the correct answer.</span>
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An object is formed by attaching a uniform, thin rod with a mass of mr = 6.9 kg and length L = 4.88 m to a uniform sphere with m
Bumek [7]

Answer:

a)  total moment of inertia is 1359.05 kg m^2

b) angular acceleratio is 0.854rad/sec^2

Explanation:

Given data:

m1=6.9 kg

L=4.88 m

m2=34.5 kg

R=1.22 m

we klnow that moment of inertia for rod is given as

J1=(1/12) ×m×L^2

J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2

As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R

For rod distance is  d1=0.5*L

By Steiner theorem

for the rod we get J_1'=J_1 + m_1\times d_1^2

J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2

for the sphere we get J_2' = J_2 + m_2\times d_2^2

J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2

And the total moment of inertia for the first case is

J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2

b) F=476 N

The torque for system is given as

M = F\times d\times sin(a)

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree  

M = F\times L/2

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

a_1 = \frac{M}{J_{t1}}

      = \frac{1161.44}{1359.05}

      = 0.854 rad/sec^2

4 0
3 years ago
And I need help with seven and eight only I will appreciate it
Mumz [18]

Answer:

7] Force = mass × acceleration

Force = 2 × 5

<u>Force = 10 N</u>

<u></u>

8] Velocity = acceleration due to gravity × time taken

Velocity = 9.8 × 12

<u>Velocity = 117.6 m/s</u>

8 0
2 years ago
2 kg
pochemuha

Answer:

2500 and kg its is very easy do with method

8 0
3 years ago
A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of
Paha777 [63]

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

Workdone = force * distance

Making "distance" the subject of formula, we have;

Distance = \frac {workdone}{force}

Substituting into the equation, we have;

Distance = \frac {195}{56.7}

Distance = 3.4 meters.

8 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
Oksanka [162]

Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

so B meet A 0.01 km east of flagpole

4 0
3 years ago
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