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rjkz [21]
3 years ago
13

PART 2 OF ENERGY AND FORCES UNIT TEST

Physics
1 answer:
katrin2010 [14]3 years ago
4 0

Answer:

1. at least two charged interacting parts

2. from the electric fields of charged subatomic particles

3 an arrow released from the bow

4Electrical fields of charged particles interact, bonding those with opposite charges.

5 the interaction of the electric fields of protons and electrons

6 The energy stored in the system increases.

7 Kinetic energy increases because the magnets move in the direction of the field.

8 Iron pieces accelerate toward the magnet, and the energy stored in the system decreases.

9  

The energy stored in the field decreases because the magnet moves in the direction of the field.

10 The energy stored increases and then decreases.

11 The wire was not connected to the source.

12  The electromagnet will become more powerful.

the rest are written, hope this helps (:

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A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the
Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2= 85.75 J

The speed of the ball after leaving the bat is:

K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}

V_f=42.5m/s

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