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dlinn [17]
2 years ago
11

If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g

ravitational field? (Note: The radius of the moon is roughly 1080 miles)
Physics
1 answer:
yulyashka [42]2 years ago
5 0

Answer:

178200 g mile pounds

Explanation:

Work= Force * Distance= Fh

F=ma=mg where m is mass and g is acceleration due to gravity

Work= 165 pounds *g* 1080 m=  178200 g mile pounds

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A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

\texttt{ }

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

\boxed {I = \Sigma F \times t}

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

Ek = 70\% \ Ek_o

\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )

v^2 = 70 \% \ (v_o)^2

v = - v_o \sqrt{70 \%} → <em>negative sign due to ball rebounds</em>

v = - v_o \sqrt{0.7} \texttt{ m/s}

\texttt{ }

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

\Delta p_{stone} = - \Delta p_{ball}

\Delta p_{stone} = - [ mv - mv_o ]

\Delta p_{stone} = m[ v_o - v ]

\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]

\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]

\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]

\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Average Speed of Plane : brainly.com/question/12826372
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0
2 years ago
How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?
podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J

Final velocity = 40 m/s

\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

4 0
1 year ago
In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce
noname [10]

Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

Inserting the values

v = 0 + (490) (0.030)

v = 14.7 m/s

6 0
2 years ago
Distinguish between the terms strength, power, and endurance as they are used in weightlifting ​
motikmotik

Answer:

Yes

Explanation:

Because I know

6 0
1 year ago
Read 2 more answers
Learning Goal: To practice Problem-Solving Strategy 40.1 for quantum mechanics problems. Suppose a particle of mass m is confine
satela [25.4K]

Answer

The answer and procedures of the exercise are attached in the following archives.

Explanation  

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

7 0
1 year ago
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