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patriot [66]
4 years ago
8

Choose all facts that increase the orbital velocity of a vessel around planet B. Bigger mass of planet B smaller mass of planet

B smaller mass of vessel vessel orbiting closer to planet B
Physics
1 answer:
telo118 [61]4 years ago
6 0

Answer:

- Bigger mass of planet B  

- orbiting closer to planet B

Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

G\frac{mM}{r^2}=m\frac{v^2}{r}

where

G is the gravitational constant

m is the mass of the vessel

M is the mass of the planet

r is the distance between the vessel and the centre of the planet

v is the orbital velocity of the vessel

Re-arranging the formula, we find an expression for v:

v=\sqrt{\frac{GM}{r}}

We see that:

- the bigger the mass of the planet, M, the bigger the velocity

- the bigger the distance between the vessel and the planet, r, the smaller the velocity

So, the correct choices that increase the orbital velocity are:

- Bigger mass of planet B  

- orbiting closer to planet B

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Two 5.0-cm-diameter conducting spheres are 8.0 m apart, and each carries 0.12 mC. Determine (a) the potential on each sphere, (b
blagie [28]

Answer:

Explanation:

Two spheres 10m apart

Each charge on the sphere is 0.12mC= 0.12×10^-3C

Given that the diameter is

5cm=0.05m

Then, the radius is diameter / 2

r=d/2

r=0.05/2

r=0.025m

Potential is given as

V=kq/r

k=9×10^9Nm2/C2

a. Potential on each sphere surface.

They are going to have the same value since the sphere are identical

At the surface of the sphere,

r= 0.025m

V=kq/r

V=9E9×0.12E-3/0.025

V=4.32E7Volts

V=4.32×10^7Volts

b. The electric field at the surface of each sphere will be the same since the charge are identical,

So, Electric field is given as

E=kq/r^2

At the surface

E=9E9×0.12E-3/0.025^2

E=1.728×10^8N/C

c. The potential mid way between the two sphere

The potential difference due to the first sphere

The sphere are 10m apart then the distance mid way is 5m

Then, the radius of the sphere is 0.025m

Total distance from the center is 5.025m

Then,

V=kq/r

V1=9E9×0.12E-3/5.025

V1=2.149×10^5 Volts

Potential difference due to the second sphere is the same as the first, since both are identical

Total potential is V1 +V2

V=2.149E5+2.149E5

V=4.3E5Volts

Total potential at the middle due to the two sphere is 4.3×10^5Volts

d. The potential difference between the sphere at any point is equal to the potential difference found at c

Therefore the potential difference is

V=4.3×10^5Volts

4 0
3 years ago
What is the precision for each measurement for the online simulation. What would be the precision for force
zimovet [89]

Answer:

Unit of precision for force is the Newton.

Explanation:

It is the official unit used to describe force in science and mostly abbreviated with the symbol N.

4 0
3 years ago
A car is moving at 35 mph and comes to a stop in 5 seconds.
amid [387]

Answer:

-10.267

Explanation:

Initial speed > 35 miles per hour

Final speed > 0 miles per hour

Time > 5 seconds

4 0
4 years ago
Read 2 more answers
A listener is sitting somewhere on the line between two loudspeakers that are 10 m apart. The speakers are each emitting a sine
prohojiy [21]

Answer:

57.17 Hz 114.34 Hz 285 Hz

Explanation:

The distance between the men and 1 speaker = 3.5 m

Distance between the men and second speaker = 10-3.5= 6.5 m

Here at this point there will be no sound so there will be destructive interference

Path difference \Delta x=6.5-3.5=3

We know that for destructive interference \Delta x=(2m+1)\frac{\lambda }{2}=(2m+1)\frac{v}{2f}

3=(2m+1)\frac{v}{2f}

f=(2m+1)\frac{v}{6} here v is the speed of sound in air

So for m =0

f=(2\times 0+1)\frac{343}{6}=57.17H

for m =1

f=(2\times 1+1)\frac{343}{6}=114.34Hz

for m=2

f=(2\times 2+1)\frac{343}{6}=285Hz

 

6 0
3 years ago
Many physical properties, such as force and mass, cannot be measured directly. Rather, some other physical property is measured
Salsk061 [2.6K]

Answer:

u =0.269

Explanation:

To find the coefficient of friction we know the following formula

F_{f} = uN

Where

F_{f} = Force of Friction

u = Coefficient of Friction

N = Normal Force

Thus we first find the Normal force (N). Remember that the Normal force is perpendicular to the surface, and is equal to the opposing component of Weight (W). Since the surface here is horizontal, then the Normal force will be equal to the Weight.

N = mg\\ N = (2.12)(9.8)\\ N = 20.776 N

Now we find the Force on the spring that caused the extension of 3.25cm or 0.0325m

F_{f} = ke

Where

F_{f} = Force of Friction

k = Force Constant

= extension

Hence

F_{f} = ke\\ F_{f} = (172)(0.0325)\\ F_{f} =5.59N

Now to find the coefficient of friction we use the first formula

F_{f} = uN\\ 5.59 = u(20.776)\\ u =0.269

4 0
3 years ago
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