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4 years ago

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A runner is training for an upcoming marathon by running around a 80-m-diameter circular track at constant speed. Let a coordina

te system have its origin at the center of the circle with the x-axis pointing east and the y-axis north. The runner starts at (x, y) = (48 m, 0 m) and runs 2.5 times around the track in a clockwise direction.1. What is the magnitude of his displacement vector?2. What is the direction of his displacement vector?

1 answer:

Umnica [9.8K]4 years ago

7 0

Answer:

Explanation:

Given

diameter $d=80\ m$

runner start at (x,y)=(48 m,0 m)

Runner runs around the circle 2.5 times

i.e. he runs two and half rounds

displacement of two rounds is zero so net displacement is of half round

displacement $=\pi \cdot 48=150.816\ m$

After 2.5 round he will be towards west direction with respect to origin

so direction of displacement vector is west

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Given that the mass of the raindrop is

$m=7.7\times10^{-7}\text{ kg}$

The acceleration due to gravity is g = 9.81 m/s^2

We have to find

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(b)Magnitude of the gravitational force exerted on the earth by the raindrop

(a) The formula to calculate the magnitude of the gravitational force exerted on the raindrop by the earth is

$F=mg$

Substituting the values, the gravitational force exerted on the raindrop by the earth is

$\begin{gathered} F=7.7\times10^{-7}\times9.81 \\ =7.55\times10^{-6}\text{ N} \end{gathered}$

(b) According to Newton's third law,

If F' is the gravitational force exerted on the earth by the raindrop, then

$F=-F^{\prime}$

Here, the negative sign indicates that both forces act in opposite direction.

The gravitational force exerted on the earth by the raindrop is

$F^{\prime}\text{ = -7.55}\times10^{-6}\text{ N}$

And the magnitude of the gravitational force exerted on the earth by the raindrop is

$7.55\times10^{-6}\text{ N}$

Thus, the magnitude of both forces is equal.

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1 year ago

The velocity of a free falling ball at different time intervals on unknown planet is shown in Figure 1.24.

dsp73

Answer:

a1 = (V2 - V1) / 1 s = 3.75 m/s^2

a2 = (V2 - V1) / (t2 - t1) = (7.5 - 3.75) / 1 s = 3.75 m/s^2

a3 = (V2 - V1) / (t2 - t1) = (7.5 - 3.75) / 1 s = 3.75 m/s^2

A) a = (a1 + a2 + a3) / 3 = 3.75 m/s^2

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B) after 1 sec v = 3.75 m / sec^2 * 1 sec = 3.75 m/s

after 2 sec v = v0 + a t = 3.75 m/s + 3.75 m/s^2 * 1 s = 7.5 m/s

C) The velocity increases proportionaly to the time

D) One factor depends on the variation of the acceleration due to gravity because of different positions of measurement (on earth)

A much larger factor is due to air resisance of the air on the falling object

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2 years ago

A coil has a resistance of 5Ω and an inductance of 100 mH . At a particular instant in time after a battery is connected across

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I have seen this question before and the correct answer would be B

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3 years ago

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3 years ago

A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn

EastWind [94]

(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, $C=5.00 \mu F$ , the charged stored on the capacitor at the end of the process is

$Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C$

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

$V=\frac{Q}{C}$

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

$C=\frac{\epsilon_0 \epsilon_r A}{d}$

where:

$\epsilon_0$ is the permittivity of free space

$\epsilon_r$ is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: $C'=\frac{C}{2}$ . The new voltage across the plate is given by

$V'=\frac{Q}{C'}$

and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{C/2}=2 \frac{Q}{C}=2V$

So, the new voltage is

$V'=2 (12.0 V)=24.0 V$

(c) (ii) 3.0 V

The area of each plate of the capacitor is given by:

$A=\pi r^2$

where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be

$A'=\pi (2r)^2 = 4 \pi r^2 = 4A$

While the separation between the plate was unchanged (d); so, the new capacitance will be

$C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C$

So, the capacitance has increased by a factor 4; therefore, the new voltage is

$V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}$

which means

$V'=\frac{12.0 V}{4}=3.0 V$

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3 years ago