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klasskru [66]
3 years ago
15

Two objects gravitationally attract with a force of 70N. If the masses of both of the objects are increased by 15x and the dista

nce between their centers change by a factor of 12x, then the new force of attraction is ___ N.
a. 1.56 N
b. 109.38 N
c. 7.29 N
d. 1312.5 N
Physics
1 answer:
erastova [34]3 years ago
7 0
(B is the answer!!
I took the test and I got a 100! Hope this helps
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iren2701 [21]

Not sure but i will say D

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If a satellite weighs 321 lb. on the earth's surface (R = 4,000 miles), how much does it weigh 12,000 miles above the surface? (
Sever21 [200]

The gravitational force between the Earth and the satellite (its "weight") is inversely proportional to the distance between the centers of both objects.

On the surface, their centers are separated by 1 Earth radius.

12,000 miles above the surface, they're separated by 4 Earth radiii.

(4/1) = 4

So after the move, the satellite's weight is (1/4²) = 1/16 of its surface weight.

(321 lb) / (16) = (20 and a hair) lb

The correct choice from the given list is " <em>>20 lb "</em> .  

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Question 11 of 15<br> what has an atomic number of 1?
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Atomic number 1,symbol H,name hydrogen
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3 years ago
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satellite is placed in an elongated elliptical (not circular) orbit around the Earth. At the point in its orbitwhere it is close
GrogVix [38]

Answer:

a) v2=4147.72 m/s

b) stotal=5.53x10^6 m

Explanation:

a) the length from the center of the earth is equal to:

L1=1x10^6+((6.37/2)x10^6)=4.18x10^6 m

the velocity is 5.14 km/s=5.14x10^3 m/s

the farthest distance is equal to:

L2=2x10^6+((6.37/2)x10^6)=5.18x10^6 m

As the angular momentum is conserved, we have to:

I1=I2

m*L1*v1=m*L2*V2, where m is the mass of satelite

clearing v2:

v2=(L1*V1)/L2=(4.18x10^6*5.14x10^3)/5.18x10^6=4147.72 m/s

b) Using the Newton 3rd law:

vf^2=vi^2+2as

where:

a=g=9.8 m/s^2

vf=0

vi=5.14 km/s

s=?

Clearing s:

s=(vf^2-vi^2)/(2g)=((0-(5.14x10^3)^2)/(2*9.8)=1.35x10^6 m

the total distance is equal to:

stotal=s+L1=1.35x10^6+4.18x10^6=5.53x10^6 m

8 0
3 years ago
Two ice skaters, Daniel (mass 70.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while a
wel

Answer:

a) v=7.32m/s

b) \alpha =-35º

c) ΔK=-1094.62J

Explanation:

From the exercise we know that the collision between Daniel and Rebecca is elastic which means they do not stick together

So, If we analyze the collision we got

p_{1x}=p_{2x}

To simplify the problem, lets name D for Daniel and R for Rebecca

a) p_{D1x}+p_{R1x}=p_{D2x}+p_{R2x}

Since Daniel's initial velocity is 0

m_{R}v_{Rx}=m_{D}v_{D2x}+m_{R}v_{R2x}

v_{D2x}=\frac{m_{R}*v_{R1x}-m_{R}*v_{R2x}}{m_{D}}

v_{D2x}=\frac{(45kg)(14m/s)-(45kg)(8cos(55.1)m/s)}{(70kg)}=6m/s

Now, lets analyze the movement in the vertical direction

p_{1y}=p_{2y}

Since p_{1y}=0

0=m_{D}v_{D2y}+m_{R}v_{R2y}

v_{D2y}=-\frac{m_{R}v_{2Ry}}{m_{D}}=-\frac{(45kg)(8sin(55.1)m/s)}{(70kg)}=-4.21m/s

Now, we can find the magnitude of Daniel's velocity after de collision

v_{D}=\sqrt{(6m/s)^2+(-4.21m/s)^2}=7.32m/s

b) To know whats the direction of Daniel's velocity we need to solve the arctan of the angle

\alpha =tan^{-1}(\frac{v_{y}}{v_{x}})=tan^{-1}(\frac{-4.21}{6})=-35º

c) The change in the total kinetic energy is:

ΔK=K_{2}-K_{1}

ΔK=\frac{1}{2}[(45kg)(8m/s)^2+(70kg)(7.32m/s)^2-(45kg)(14m/s)^2]=-1094.62J

That means that the kinetic energy decreases

5 0
3 years ago
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