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Vinvika [58]
3 years ago
7

A 5510 kg space probe, moving nose-first toward Jupiter at 250 m/s relative to the Sun, fires its rocket engine, ejecting 75.0 k

g of exhaust at a speed of 241 m/s relative to the space probe. What is the final velocity of the probe
Physics
1 answer:
kherson [118]3 years ago
5 0

Answer:

final velocity of the probe 253.30 m/s

Explanation:

Mass of space probe mi= 5510 kg

initial speed of probe vi= 250 m/s

Exhaust mass m= 75 kg

speed of exhaust vf = 241 m/s

Mass of the fuel mf= mi-m = 5510-75 =5435 kg

Final speed of probe

v_{f}=v_{i}+v_{r}ln(\frac{m_{i}}{m_{f}} )\\v_{f}=250+241 \times  ln(\frac{5510}{5435} \\v_{f}=250+3.30\\v_{f}=253.30 m/s

final velocity of the probe 253.30 m/s

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Which of your groups should you NOT change anything for? (in other
ella [17]

Answer:

B

Explanation:

The control is something that is meant to not be changed, the control is a comparison of the experimental.

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3 years ago
Please Help On These 2 Questions!!!!! I severely need help!!!!!!
lys-0071 [83]
3. In a uniform electric field, the equation for the magnitude of the magnetic field is E=(V/d). V= voltage d= distance. If the magnetic field magnitude is constant , as stated in your problem, then the voltage must stay the same otherwise the value of "E" would change". And the problem already told us the "E" is uniform and so, not changing. Does that make sense?

4a. If the magnetic field lines are equally spaced apart, in other words share the same density. Then we know that the magnitude of the magnetic field is unchanging. This is because the density of of the magnetic field lines(how many are in a certain area) is related to the magnitude being expressed by the electric field. Greater magnitude is expressed by the presence of more lines (higher line density) 

4b. The electric potential is measured in Volts(V) and is uniform along  the same equipotential line. What is an equipotential line(gray)? It is a line drawn perpendicular(forms a right angle with) to the magnetic field lines(black) to show the changes in electric potential. One space where electric potential will always be the same because it will always be equal to 0 Volts is exactly in between a positive and negative charges of equal charge value I have pointed to this line with a purple arrow in my picture.

I really hope this makes sense to you and that my pictures help! :)

3 0
2 years ago
At which temperature does the motion of atoms and molecules stop?<br> 0°C<br> 0C<br> 0°K<br> 0K
anastassius [24]

Answer: 0K

Explanation:

Absolute 0 (0K) is the point where nothing could be colder and no heat energy remains in a substance.

7 0
3 years ago
You comb your hair and the comb becomes negatively charged. Strictly speaking, how will the mass of your hair change? A. it wil
Step2247 [10]

Answer:

C. it will not change.

Explanation:

While combing, the rubbing of the comb with the hair, transfer of electron takes place from the hair to the comb and the comb becomes negatively charged. But, this transfer of electron does not make any considerable change in the mass of the hair. This is because the mass of an electron is highly negligible. Now, neglecting the mass of an electron, the transfer of the electrons from the hair to the comb makes charging of the comb, but no loss of mass in the hair. So, the mass of hair will no change.

7 0
3 years ago
A carnival Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute. What is the accelerat
Tanzania [10]

If a Ferris wheel has a 15-m radius and completes five turns about its horizontal axis every minute then the acceleration of a passenger at his lowest point during the ride is 4.11 \text{m/s}^{2}.

Calculation:

Step-1:

It is given that the radius of the Ferris wheel is r=15 m, and the angular speed of the wheel is \omega=5rev/min.

It is required to find the angular acceleration of a passenger at his lowest point during the ride.

The formula required to calculate the angular acceleration is, a=\omega ^2 r.

Step-2:

Now substituting the given values into the equation to get the value of the angular acceleration.

\begin{aligned}a &=\omega^{2} r \\&=(5 \mathrm{rev} / \mathrm{min})^{2}(15 \mathrm{~m}) \\&=\left(5 \mathrm{rev} / \mathrm{min} \times \frac{\left(\frac{2 \pi}{60}\right) \mathrm{rad} / \mathrm{sec}}{1 \mathrm{rev} / \mathrm{min}}\right)^{2}(15 \mathrm{~m}) \\&=(0.2741 \times 15) \mathrm{m} / \mathrm{s}^{2} \\&=4.11 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}

The acceleration is towards upwards that means towards the center of the wheel.

Learn more about the angular acceleration:

brainly.com/question/1592013

#SPJ4

3 0
1 year ago
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