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Montano1993 [528]
3 years ago
6

In which situation is the maximum possible work done?

Physics
1 answer:
ExtremeBDS [4]3 years ago
3 0

Answer:

Answer is A.

Explanation:

Remember the equation for work:

W = F d cos θ

Math explanation: When θ = 0, cos θ = 1, so Fd is a maximum. For any other value of θ up to 90 degrees, cos θ < 1 so Fd will be smaller. For θ = 180, you're pushing away from the direction of displacement so you're actually doing negative work.

Intuitive explanation: you have the most success pushing an object when you apply the force directly against it compared to if the force was directed at an angle.

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A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upwa
Ksju [112]

The toy rocket is launched vertically from ground level, at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance.

The total energy of the rocket, which is a sum of its kinetic energy and potential energy, is constant.

At a height of 72 m with the rocket moving at 30 m/s, the total energy is m*9.8*72 + (1/2)*m*30^2 where m is the mass of the rocket.

At ground level, the total energy is 0*m*9.8 + (1/2)*m*v^2.

Equating the two gives: m*9.8*72 + (1/2)*m*30^2 = 0*m*9.8 + (1/2)*m*v^2

=> 9.8*72 + (1/2)*30^2 = (1/2)*v^2

=> v^2 = 11556/5

=> v = 48.07

<span>The velocity of the rocket when it impacts the ground is 48.07 m/s</span>

3 0
3 years ago
What the effect positively charged objects and negatively charged objects have on each other
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3 years ago
To test a slide at an amusement park, a block of wood with mass 3.00 kgkg is released at the top of the slide and slides down to
dem82 [27]

Answer:

511.1 J

Explanation:

We are given that

Mass of wood block=m=3 kg

Vertical distance,h=23 m

Horizontal distance =x=30 m

Distance traveled in downward direction y=40 m

Initial velocity,u=0

y=ut+\frac{1}{2} gt^2

Where g=9.8 m/s^2

40=0+\frac{1}{2}(9.8)t^2=4.9t^2

t^2=\frac{40}{4.9}

t=\sqrt{\frac{40}{4.9}}=2.86 s

x=v_x\times t

30=v_x(2.86)

v=v_x=\frac{30}{2.86}=10.49 m/s

By work energy theorem

Change in kinetic energy=Work done= mgh-W

\frac{1}{2}mv^2-\frac{1}{2}mu^2=3\times 9.8\times 23-W

\frac{1}{2}(3)(10.49)^2-0=676.2-W

W=676.2-\frac{1}{2}(3)(10.49)^2=511.1 J

Hence, the work done due to friction on the block as it slides down the ramp=511.1 J

6 0
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