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Amanda [17]
2 years ago
12

A dog runs 300m North and sees the dog catcher and runs 120 m south .Whats the dogs displacement.The same jogger runs 3 miles ar

ound the track, starting and finishing in the same time after completing exactly 12 laps. What is the average velocity now
Physics
1 answer:
ozzi2 years ago
8 0

Answer:

<em>1. 180 m</em>

<em>2. 1.5 miles/hour</em>

<em>3. 0 m/s </em>

Explanation:

The complete question is

A dog runs 300 m North and sees the dog catcher and runs 120 m south Whats the dogs displacement.

A jogger runs north for 3.0 miles. If this took 2.0 hours, what is the jogger’s average velocity in miles per hour?

The same jogger runs 3.0 miles on the track, starting and finishing his run after completing exactly 12 laps around the track. What is his average velocity?

1. Displacement is how far the dog moves from the original starting point.

If the dog runs 300 m north, and then 120 m south, then the displacement from the starting point will be

displacement = 300 - 120 = <em>180 m</em>

2. Displacement of the jogger = 3 miles

Time taken by the jogger = 2 hours

velocity of the jogger = ?

Velocity is = displacement/time

velocity = 3/2 = <em>1.5 miles/hour</em>

3. velocity = displacement/time

if the jogger completes exactly 12 laps around the track, then the jogger will return to the starting point. This means that the jogger's displacement from the starting point is 0 miles

Jogger's velocity will therefore be = <em>0 m/s </em>

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Answer:

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Explanation:

The equation for force is given:

F=m*a

First, we must find the total mass, which is the sum of the boy's mass and the go-cart's mass.

  • total mass= boy's mass + go cart's mass

The boy's mass is 35 kilograms and the go cart's is 65 kilograms.

  • total mass= 35 kg+ 65 kg=100 kg

Now we know the total mass and the acceleration.

m= 100 \ kg \\a= 5 \ m/s^2

Substitute the values into the formula.

F=100 \ kg * 5 \ m/s^2

Multiply.

F= 500 \ kg*m/s^2

  • 1 kilograms meter per square second is equal to 1 Newton.
  • Our answer of 500 kg*m/s² is equal to 500 Newtons.

F= 500 \ N

The force exerted by the go cart engine is <u>500 Newtons.</u>

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A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem
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Answer:

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The energy dissipated in the resistor {U_k} = \frac{U}{k}

B

The energy dissipated in the resistor{U_k} = kU

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

                 U _k = \frac{1}{2} \frac{q ^2}{kC}

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

​ U= \frac{1}{2} \frac{q ^2}{C}

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation q = CV is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2}  kCV ^2

In this equation, Uk is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

 U _k = \frac{1}{2} \frac{q ^2}{kC}

Substitute U = \frac{1}{2}\frac{{{q^2}}}{C}  in the equation of {U_k}

U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

U_{k} = \frac{1}{2}kCV^2

Here, V is voltage in the circuit.

Substitute U =\frac{1}{2} CV^2 in the equation of {U_k}

So,

        U_{k} = \frac{1}{2} kCV^2\\

       = k(\frac{1}{2} CV^2)

       U_{k} = kU

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Answer:

They will come back at the same time.

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I hope this answer helps.

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