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stich3 [128]
2 years ago
13

Can a constant magnetic field set into motion an electron initially at rest? Explain your answer.

Physics
1 answer:
Vinvika [58]2 years ago
5 0

No,  a constant magnetic field cannot set an electron initially at rest into motion

A force that accelerates a particle is necessary to change its velocity. The magnetic force is inversely proportional to the particle's speed. There cannot be a magnetic force acting on a moving particle, according to Einstein. A flux is a precise description of the greater-than-unity magnetic determine involving energy currents and magnet resources. The magnetic flux in a stage is actually selected apart from each some sort of route and also a degree (or durability); therefore, it is just a vector industry. The magnetic flux is usually defined as the Lorentz force that acts on moving galvanic costs.

To know more about Lorentz force refer to brainly.com/question/15552911

#SPJ4

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In the circuit shown, the galvanometer shows zero current. The value of resistance R is :
Tasya [4]

Answer:

its supposed to be (a) 1W

8 0
3 years ago
Diane swim 5 kilometers against the current in the same amout of time it took her to swim 15 kilometers with the current . The r
lidiya [134]

Answer:

4 km/hr

Explanation:

suppose 's' is Diane's speed with no current.

't' represents time in hrs.

Using the formula:

Distance = speed 's' x time 't'

-> when she swims against the current, equation will be,

5= (s-2)t

t= 5/(s-2)

->when she was swimming with the current, equation is,

15= (s+2) t

t= 15/(s+2)

equating eq(1) and (2)

5/(s-2) = 15/(s+2)

5s + 10 = 15s - 30

40= 10s

s= 40/10

s=4

Therefore, if there were no current, her speed is 4km/hr

5 0
4 years ago
You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of 7.00 m above the c
Mariana [72]

Answer:

Your friend has to wait 0.26 s after you throw the ball to start running.

Explanation:

The equation that gives the position vector of the ball is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal positon

v0 = initial velocity

t = time

α = throwing angle

y0 = initial vertical position

g = acceleration due to gravity

The equation of displacement of your friend is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of your friend at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.

Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:

y = y0 + v0 · t ·sin α + 1/2 · g · t²

-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m

Solving the quadratic equation:

t = 1.71 s

Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):

x= x0 + v0 · t · cos α

x = 0m + 9.00 m/s · 1.71 s · cos 33°

x = 12.9 m

Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.

Let´s see, how much time it takes your friend to run that distance:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0, v0 = 0)

x = 1/2 · a · t²

1.9 m = 1/2 · 1.80 m/s² · t²

Solving for t

t = 1.45 s

Then, since the time of flight of the ball is 1.71 s, your friend has to wait

1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.

6 0
3 years ago
A child\'s top is held in place, upright on a frictionless surface. The axle has a radius of r = 2.96 mm. Two strings are wrappe
sladkih [1.3K]

Answer:

Explanation:

Given

radius r=2.96 mm

Tension T=2.4 N

time taken=0.74 s

Let \alphabe the angular acceleration

2 T\times r=I\times \alpha

2\times 2.4\times 2.96\times 10^{-3}=0.5\times m\times (2.96\times 10^{-3})^2\times \alpha

\alpha =\frac{4\times 2.4}{m\times 2.96\times 10^{-3}}

\alpha =\frac{3.24\times 10^3}{m} rad/s^2

\omega =\omega _0+\alpha \cdot t

\omega =0+\frac{3.24\times 10^3}{m}\times 0.74

\omega =\frac{2.4\times 10^3}{m} rad/s

Angular momentum

L=I\omega

L=0.5\times mr^2\times \omega

L=0.5\times m\times (2.96\times 10^{-3})^2\times \frac{2.4\times 10^3}{m}

L=0.01051 kg-m^2/s

4 0
3 years ago
8. What is the rate of change in an object’s position?
adelina 88 [10]
A similar but separate notion is that of velocity, which the rate of change<span> of </span>position<span>. Example . If p(t) is the </span>position<span> of an </span>object<span> moving on a number line at time t (measured in minutes, say), then the average </span>rate of change<span> of p(t) is the average velocity of the </span>object<span>, measured in units per minute.</span>
4 0
3 years ago
Read 2 more answers
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