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anastassius [24]
3 years ago
11

1. A rocket hits the ground at an angle of 60° from the horizontal at a speed of 300 m/s.

Physics
1 answer:
alekssr [168]3 years ago
5 0

Answer:

(a). The horizontal and vertical components are v\cos\theta and v\sin\theta

(b). The horizontal and vertical components of the rocket's impact velocity is 150 m/s and 259.8 m/s.

Explanation:

Given that,

Angle = 60°

Speed = 300 m/s

(a). We need to draw the vector representing the rocket's impact and its components

Using given data

The components of rocket's impact

The horizontal component is

u_{x}=u\cos\theta

The vertical component is

v_{v}=v\sin\theta

(b). We need to calculate the horizontal and vertical components of the rocket's impact velocity

Using horizontal and vertical components

The horizontal component is

v_{x}=v\cos\theta

Put the value into the formula

v_{x}=300\cos60

v_{x}=150\ m/s

The vertical component is

v_{x}=v\sin\theta

Put the value into the formula

v_{x}=300\sin60

v_{x}=259.8\ m/s

Hence, (a). The horizontal and vertical components are v\cos\theta and v\sin\theta

(b). The horizontal and vertical components of the rocket's impact velocity is 150 m/s and 259.8 m/s.

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A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
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Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

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Answer:

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W = 0

Explanation:

Work is said to be done whenever a force of one newton moves a body of one kilogram through a distance of one meter. Meaning the applied force has to move the body from a point of rest through certain distance.

Work = force × distance

So, in the case of this question, we only have the force been applied, but no distance was covered. Hence, no work was done.

W = 3000× 0 meter

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