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harina [27]
3 years ago
15

A certain planet has an escape speed . If another planet has twice size and twice the mass of the first planet, its escape speed

will be
A) Sqrt[2] V
B) V
C) V/2
D) V/Sqrt[2]
E) 2V
Physics
2 answers:
LenKa [72]3 years ago
7 0

Answer:

answer is V

Explanation:

solong [7]3 years ago
3 0
I think the correct answer from the choices listed above is option A. A certain planet has an escape speed . If another planet has twice size and twice the mass of the first planet, its escape speed will be <span>Sqrt[2] V. Hope this answers the question.</span>
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You are spinning a rock, of mass 0.75 kg, at the end of a string of length 0.86m in a vertical circle in uniform circular motion
Nina [5.8K]

Answer:

v (minimum speed) = 2.90 m/sec.

\\ \\ maximum speed (v)= 6.57 m/sec.\\

Maximum value of speed will occur at lowest point of vertical circle.

Explanation:

a)  What minimum speed is necessary so that there is no tension in the string at the top of the circle but the rock stays in the same circular path?

Using the force balance expression at the top of the circle,

Gravitational Force + Tension force = Centrifugal force

m*g + T = m*v^2/R

Given that : T = 0

R = length of string = 0.86 m

mass of the spinning rock = 0.75 kg

v = \sqrt{g*R}

v = \sqrt{9.81*0.86}

v (minimum speed) = 2.90 m/sec.

b) what is the maximum speed the rock can have so that the string does not break?

Here the  force balance at bottom of circle is represented by the illustration:

T = m*g + m*v^2/R

Given that:

maximum tension T = 45 N

maximum speed v = ??

mass  m = 0.75 kg

∴

45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\

c)

At what point in the vertical circle does this maximum value occur?

Maximum value of speed will occur at lowest point of vertical circle.

This is so because  at the lowest point; the tension in string will be maximum.

4 0
3 years ago
Explica de que tipo es cada oración según la actitud del hablante
Volgvan

Answer:

Según la actitud del hablante las oraciones se clasifican en enunciativas, interrogativas, etc. ... adverbios o expresiones que complementan a toda la oración (COr): ojalá, quizá.

Explanation:

7 0
2 years ago
a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
lilavasa [31]

Answer:

D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

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2 years ago
How do the tension of the cord and the force of gravity affect a pendulum?
LuckyWell [14K]

Answer:

<em>Force of gravity may not affect a pendulum during its equilibrium state</em>. But  the gravity can affect the pendulum when a force occurs in any direction of the bob connected to the cord that makes a swing sideways. The gravity of pendulum never stops, it always accelerates. So the gravity affects the pendulum acceleration and speed.    

<em>Similarly the tension in the cord will not affect the pendulum</em><em> </em>but if change in the length of the pendulum while keeping other factors constant changes the length of the period of pendulum. longer pendulum swings with lower frequency than shorter pendulums.    


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3 years ago
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Marat540 [252]

Answer:

The answer are given above in attachment.

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