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2 years ago

If you run around a circular track with a radius of 10m, what is the displacement after 3 full laps back to the start line?

Physics

1 answer:

Svetach [21]2 years ago

8 0

Displacement will be zero.

<h3>What is displacement and how is it different from distance?</h3>

Displacement:

It is the shortest distance between an object's beginning position and ending position.
The quantity is a vector.
It may be zero, positive, or negative.
Displacement may be greater than or equal to distance.

In the given question, we start from one point and finish at the same point around the circular track. Therefore there is no distance between the initial and final points because they are the same. That's why the displacement is zero.

Distance:

It is the actual path or distance that an object takes to go from its starting point to its finishing point.
The quantity is a scalar one.
It cannot be 0 or negative; it must be positive.
Displacement and distance may be equal.

Learn more about displacement here:

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You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe

steposvetlana [31]

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, $L=800\ mH=800\times 10^{-3}\ H=0.8\ H$

Capacitance, $C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F$

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

$Z=\sqrt{R^2+(X_C-X_L)^2}$ ...........(1)

Where

$X_C$ is the capacitive reactance, $X_C=\dfrac{1}{2\pi fC}$

$X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega$

$X_L$ is the inductive reactance, $X_L={2\pi fL}$

$X_L={2\pi \times 60\times 0.8}=301.59\ \Omega$

So, equation (1) becomes :

$Z=\sqrt{(100)^2+(265.65-301.59)^2}$

Z = 106.26 ohms

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.

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3 years ago

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Answer:

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1 year ago

Does heat demagnetize?

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Answer:

Yes

Explanation:

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3 years ago

An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a

Liono4ka [1.6K]

Answer:

$16.25^{\circ}$

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Horizontal range is given by

$R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}$

The angle at which the arrow is to be released is $16.25^{\circ}$ .

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