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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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PLEASE ANSWER ASAP!!! ITS DUE IN 10 MINUTES!!!!!!
forsale [732]

Explanation:

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Strength building. ...

Balance Training. ...

Endurance. ...

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Vigorous exercise.

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Hope it helps..

5 0
4 years ago
A 36,287 kg truck has a momentum of 907,175 kg • . What is the truck’s velocity?
Snowcat [4.5K]
By definition,
Momentum = Mass * Velocity

Let v =  the velocity of the truck, m/s
The mass of the truck is 36,287 kg.
The momentum is 907,175 (kg-m)/s.

Therefore
907,175 (kg-m)/s = (36287 kg)*(v m/s)
v = 907175/36287 = 25 m/s

Answer: 25 m/s

7 0
4 years ago
Read 2 more answers
a ball of mass 15 kg moving at +20 m/s collides head-on with another ball of mass 10 kg moving at -15 Ms. if after collision the
Karo-lina-s [1.5K]

Given

m1 = 15kg

vi1 = +20 m/s

vf1 = +5 m/s

m2 = 10 kg

vi2= -15 m/s

vf2 = ?

Procedure

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant.

\begin{gathered} m_1v_{i1}+m_2v_{i2}=m_1v_{f2}+m_2v_{f2} \\ 15kg\cdot20m/s+10\operatorname{kg}\cdot-15m/s=15\operatorname{kg}\cdot5m/s+10\operatorname{kg}\cdot v_{f2} \\ 150-75=10v_{f2} \\ v_{f2}=75/10 \\ v_{f2}=7.5\text{ m/s} \end{gathered}

The speed of the 10kg Ball is 7.5m/s

8 0
1 year ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
Please select the word from the list that best fits the definition I study French so I can talk to my cousin who lives in Paris
salantis [7]

Answer:

There is no list...

Explanation:

Post the list and words so I can help you.

8 0
3 years ago
Read 2 more answers
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