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4 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.

2 answers:

goblinko [34]4 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]4 years ago

7 0

Answer:

8.0

Explanation:

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2 years ago

A spring has a 12 cm length. When a 200-g mass is hung from the spring, it extends to 27 cm. The hanging mass were pulled downwa

BartSMP [9]

Answer:

The equation of the time-dependent function of the position is $x(t)=5\cos(8.08t)$

(b) is correct option.

Explanation:

Given that,

Length = 12 cm

Mass = 200 g

Extend distance = 27 cm

Distance = 5 cm

Phase angle =0°

We need to calculate the spring constant

Using formula of restoring force

$F=kx$

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{200\times10^{-3}\times9.8}{(27-12)\times10^{2}}$

$k=13.06\ N/m$

We need to calculate the time period

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{0.2}{13.6}}$

$T=0.777\ sec$

At t = 0, the maximum displacement was 5 cm

So, The equation of the time-dependent function of the position

$x(t)=A\cos(\omega t)$

Put the value into the formula

$x(t)=5\cos(2\pi\times f\times t)$

$x(t)=5\cos(2\pi\times\dfrac{1}{T}\times t)$

$x(t)=5\cos(2\pi\times\dfrac{1}{0.777}\times t)$

$x(t)=5\cos(8.08t)$

Hence, The equation of the time-dependent function of the position is $x(t)=5\cos(8.08t)$

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3 years ago

A 900 N crate is being pulled across a level floor by a force F of 340 N at an angle of 23° above the horizontal. The coefficien

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Answer:

0.958891203 m/s²

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Force of friction acting on the force applied

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Force used to pull the crate

$F=340cos 23\\\Rightarrow F=312.97165\ N$

The net force is

$F_n=F-f\\\Rightarrow F_n=312.97167-225\\\Rightarrow F_n=87.97167\ N$

Acceleration is given by

$a=\frac{F_n}{m}\\\Rightarrow a=\dfrac{87.97167}{\dfrac{900}{9.81}}\\\Rightarrow a=0.958891203\ m/s^2$

The magnitude of the acceleration of the crate is 0.958891203 m/s²

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4 years ago

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puteri [66]

Answer:

Explanation:

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m v₂ = 2m v sin55.08

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m x 13 = 2m v cos55.08

Dividing these two equations

v₂ / 13 = tan55.08

v₂ = 13 tan55.08

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= (18.62 x60 x 60) / 1000

= 67 km/h

= 67 x 5/8 mi/h

= 42 mi/h

So he is lieing.

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scoray [572]

Answer:

read the explanation

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3 years ago