If you are measuring how far a ball traveled in an experiment which example below could be an

Which statement best defines why sound waves are mechanical waves?

Annette [7]

The correct answer among the choices presented above is option A. Sound waves are considered as mechanical waves because they require a medium in which they can travel through. A mechanical wave is a wave that cannot transmit energy in a vacuum. It needs something to move energy.<span />

7 0

2 years ago

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A long solenoid has 103 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.60 cm perp

Aneli [31]

Answer:

0.23348 A

Explanation:

B = Magnetic field

v = Velocity of electron = $1.38\times 10^7\ m/s$

q = Charge of electron = $1.6\times 10^{-19}\ C$

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

r = Radius of circle = 0.026 m

N = Number of turns = 103 turns/cm = $103\times 100\ turns/m$

I = Current

The magnetic and centripetal force will be balanced

$Bqv=m\frac{v^2}{r}\\\Rightarrow B=\frac{mv}{qr}$

The magnetic field in solenoid is given by

$B=N\mu_0 I\\\Rightarrow I=\frac{B}{N\mu_0}$

From the first equation

$I=\frac{\frac{mv}{qr}}{N\mu_0}\\\Rightarrow I=\frac{mv}{N\mu_0qr}\\\Rightarrow I=\frac{9.11\times 10^{-31}\times 1.38\times 10^7}{103\times 100\times 4\pi \times 10^{-7}\times 1.6\times 10^{-19}\times 0.026}\\\Rightarrow I=0.23348\ A$

The current in the solenoid is 0.23348 A

8 0

2 years ago

Need help asap please......

alexandr402 [8]

Answer:

The Answer is A. The slope is upward.

Explanation:

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2 years ago

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If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?

Lena [83]

Answer:2 volts

Explanation:

Power=current x voltage

6=3 x voltage

Divide both sides by 3

6/3=(3 x voltage)/3

2=voltage

Voltage=2volts

5 0

3 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

4 0

2 years ago