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Katarina [22]
2 years ago
11

The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif

ference between the two plates The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential difference between the two plates _______.a. Increases. b. Decreases. c. Does not change.
Physics
1 answer:
Dovator [93]2 years ago
3 0
The answer is B, because it will lose potential energy.
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alex41 [277]

Yeah, hand signals are important in officiating games. The hand signals are given by the referee who first signals the fault and then indicates which team has won the point. A point is indicated by one finger at the side of the court to indicate the winner of the rally.

5 0
2 years ago
Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in
Elanso [62]

Answer:

kinetic energy

Explanation:

a certain amount of energy is transferred by the kick. The ball gains an equal amount of energy, mostly in the form of kinetic energy.

4 0
2 years ago
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3. Christina is on her college softball team, and she is practicing swinging the bat. Her coach wants her to work on the speed a
inessss [21]

Answer:

The average angular acceleration is  \alpha =125.487 rad /s^2

Explanation:

From the question we are told that

  From the question we are told that

        The length of the bat is l = 0.85m  \

         The initial linear velocity is  u = 0 m/s

         The time is  t = 0.15s

         The velocity at t is  v = 16 m/s

  Generally average  angular acceleration is mathematically represented as

                \alpha  = \frac{w_f - w_o}{t}

        Where w_f is the finial angular velocity which is mathematically evaluated as  

            w_f = \frac{v}{l}

                  w_f = \frac{16}{0.85}

                        = 18.823 rad/s

 and w_o is the initial angular velocity which is zero since initial linear velocity is zero

               So

                         \alpha  = \frac{18.823 - 0}{0.15}

                               \alpha =125.487 rad /s^2

5 0
3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

E=Pt=(9.36)(350)=3276 J

Finally, the change in temperature of an object is related to the energy supplied by

E=C\Delta T

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

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