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3 years ago

7

what happens if a voltmeter is connected in series with other components of the circuit (i.e , ammeter, cell, battery, resistor

and wires). Why is it so?

1 answer:

Talja [164]3 years ago

8 0

Answer:the voltmeter measures the potential difference of the circuit,. Voltmeter is a device used to measure potential difference.

Explanation:

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Find the wavelength λ of the 80.0-khz wave emitted by the bat. express your answer in millimeters.

vfiekz [6]

Answer:

4.29 millimeters

Explanation:

Bats emit ultrasound waves: in air, ultrasound waves travel at a speed of

$v=343 m/s$

The frequency of the waves emitted by this bat is:

$f=80.0 kHz = 80,000 Hz$

Therefore we can find the wavelength of the wave emitted by the bat by using the relationship between speed, frequency and wavelength:

$\lambda=\frac{v}{f}=\frac{343 m/s}{80,000 Hz}=4.29\cdot 10^{-3} m=4.29 mm$

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2 years ago

What is the definition of the coefficient of kinetic friction?

Nookie1986 [14]

The coefficient of kinetic friction<span> is the force between two objects when one object is moving, or if two objects are moving against each other</span>

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3 years ago

True or false velocity is the slope of the acceleration versus time graph

Elena-2011 [213]

I'm pretty sure it is true. (78% sure)

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2 years ago

Read 2 more answers

If you wanted the pitch of a horn to drop relative to an observer, which way would you move the horn, relative to where that obs

Vladimir [108]

We assume that horn releases sound of constant frequency. In order for observer to observe different frequency either horn or observer or both must move.

This happens due to Doppler effect. It states that when position of source of sound and observer relative to each other changes, the observed frequency also changes. If the source emits sound of constant frequency than observed frequency will be either higher or lower than original.

When distance between source and observer increases the observed frequency will be lower. This is because same number of sound waves must cover greater distance so they have greater wavelength.
When distance between source and observer decreases the observed frequency will be higher. This is because same number of sound waves must cover smaller distance so they have smaller wavelength.

Wavelength and frequency are inversely proportional meaning when one increases the other drecreases.

From this explanation we can find answer for our question. <span>If we wanted the pitch of a horn to drop relative to an observer we need to move horn away from an observer.</span>

3 0

2 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago