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irina1246 [14]
3 years ago
6

A charged ball is moving horizontally and perpendicular to a magnetic field of 0.8 Tesla. The ball has a mass of 0.007 kg and ha

s a charge of -0.005 C. How fast must the ball be moving in order to cancel out the effect of gravity? Give the velocity as a positive number.
Physics
2 answers:
goblinko [34]3 years ago
8 0

Answer:

17.15 m/s

Explanation:

Parameters given:

Magnetic field, B = 0.8 T

Mass of ball, m = 0.007 kg

Charge of ball, q = 0.005 C

The magnetic force acting on the charged ball due to the magnetic field is given as:

F = qvBsinθ

where v = velocity of the ball and θ = angle between the horizontal and the magnetic field = 90°

The force of the ball will be in the opposite direction but of equal magnitude:

F_b = -qvBsin(90) = -qvB

To cancel out the effect of gravity, the magnetic force must be equal to the gravitational force acting on the ball:

F = mg

Therefore:

mg = -qvB

Solving for velocity, v, we have:

v = \frac{mg}{-qB}

v = \frac{0.007 * 9.8}{-(-0.005) * 0.8}

v = 17.15 m/s

The ball must be moving at a velocity of 17.15 m/s.

mezya [45]3 years ago
7 0

Answer:

8.0

Explanation:

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Answer:

R2 = 10.31Ω

Explanation:

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R1 =  13 Ω

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The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}            (2)

V: emf source voltage = 23 V

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You calculate the Req by using the equation (2):

R_{eq}=\frac{23V}{4A}=5.75\Omega

Now, you can calculate the unknown resistor R2 by using the equation (1):

\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega

hence, the resistance of the unknown resistor is 10.31Ω

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3 years ago
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dexar [7]

She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

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For a standard rectilinear lens,

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Focal length (f) and field of view (FOV) of a lens are inversely proportional.

From the equation we can say that,

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She should use shorter focal length to fit the entire landscape which she is trying to photograph into her picture.

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1 year ago
What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes
mr Goodwill [35]
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I= \frac{Q}{\Delta t}
Since I=500 A and the time interval is
\Delta t=4.0 min=240 s
the charge is
Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C

One electron has a charge of q=1.6 \cdot 10^{-19}C, therefore the number of electrons that pass a point in the wire during 4 minutes is
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3 years ago
Two bicycle tires are set rolling with the same initial speed of 4.0 m/s along a long, straight road, and the distance each trav
umka2103 [35]

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

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The acceleration of a bicycle will be:

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On substituting the given values, we get

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As we know,

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