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Oksanka [162]
3 years ago
13

A student swings back and forth from position A to C, as shown. Which of the following happens when the swing moves from Positio

n B to Position A?
[] Both potential energy and kinetic energy of the student increase.
[] Both potential energy and kinetic energy of the student decrease.
[] Potential energy of the student decreases and kinetic energy of the student increases.
[] Kinetic energy of the student decreases and potential energy of the student increases.
Physics
2 answers:
wolverine [178]3 years ago
8 0

Answer:

its B

Explanation: Its 100% verified

anygoal [31]3 years ago
6 0

Answer:[] Both potential energy and kinetic energy of the student increase.

Explanation:

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State two condition necessary for a solid to float in a liquid
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Answer:

it's density must be less than water

law of floatation

wt of the immerged body = wt of the water displaced by it

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Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Dete
podryga [215]

Answer:

Explanation:

The velocity at the inlet and exit of the control volume are same V_i=V_e=V

Calculate the inlet and exit velocity of water jet

V=V_j+V_e\\\\V=30+14\\\\V=44m/s

The conservation of mass equation of steady flow

\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0

A_i\ \texttt {is the inlet area of the jet}

A_e\ \texttt {is the exit area of the jet}

since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal

The expression for thickness of the jet

A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}

R is the radius

t is the thickness of the jet

D_j is the diameter of the inlet jet

t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm

(b)

R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)]

1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j

R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N

The negative sign indicate that the direction of the force will be in opposite direction of our assumption

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At NASA's Zero Gravity Research Facility in Cleveland, Ohio, experimental payloads fall freely from rest in an evacuated vertica
Diano4ka-milaya [45]

Answer:

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(c) velocity in mph is 115.0mph

Explanation:

(a).

The potential energy P of the payload of mass m is at a vertical distance h is  

P =mgh.

Therefore, for the payload of mass m = 50kg at a vertical distance of h = 132 m, the potential energy is

P = (50kg)(9.8m/s^2)(132m)

\boxed{P = 64,680J}

(b).

When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,

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\boxed{v = 51.43m/s}

(c).

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\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}

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How to solve for time given distance and velocity
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Answer:

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S = Vt + S0

With S being the distance, V the velocity, t the time and S0 the initial distance (initial displacement).

From this you can calculate t, if that's what you want.

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Which of these is a benefit of replacing a coal-burning power plant with a
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