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strojnjashka [21]
3 years ago
7

as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p

article has a speed of 230 m/s and moves perpendicular to the magnetic field. Just as the particle enters the magnetic field, an electric field is turned on. What must be the magnitude of the electric field such that the net force on the particle is twice the magnetic force
Physics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

E = (3)(230\ m/s)(0.61\ T)Sin90^o

<u>E = 420.9 N/C</u>

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21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest
zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

mgh=\frac{1}{2}mu^2

where the term on the left is the potential energy while the term on the right is the kinetic energy, and where

m = 50.0 kg is the mass of the youngster

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.00 m is the heigth of the platform

u is the speed of the youngster as she reaches the floor

Solving for u,

u=\sqrt{2gh}=\sqrt{2(9.8)(1.00)}=4.43 m/s

Then, when the youngster hits the floor, the force exerted on her during the deceleration is given by:

F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}

where \Delta p is her change in momentum, and where

m is the mass

v = 0 is the final velocity (she comes to a stop)

u = 4.43 m/s is the initial velocity

\Delta t=10.0 ms =0.010 s is the duration of the collision

Substituting,

F=\frac{(50.0)(0-4.43)}{0.010}=-22150 N

And the negative sign means the direction of the force is opposite to the motion (so, upward).

6 0
4 years ago
Please help! This is due tomorrow and I absolutely need help.
zysi [14]

Answer:

Correct answer:  11. Total distance d = 200m ; 12. Vav = 3.63m/s ;

13. Total displacement Dt = 0m ; 14. V₂(10s-15s) = 0 m/s ;

15. V₃(15s-40s) = 4 m/s ; 16. V₁(0s-10s) = 6 m/s > V₄(40s-55s) = 2.67 m/s

Explanation:

The whole movement can be divided into four stages.

In the first stage the subject moves 60m in a positive direction for 10s,

in the other it is stationary for 5s, in the third it moves 100m in the opposite (negative) direction for 25s and in the fourth in the positive 40m for 15s.

11. Total distance = 60 + 0 + 100 + 40 = 200m

12. The formula for calculating the average speed (velocity) is

Vav = (S₁ + S₂ + S₃ + S₄) / (t₁ + t₂ + t₃ + t₄)

Vav = (60 + 0 + 100 + 40)/ (10 + 5 + 25 + 15) = 200/55 = 3.63 m/s

13. The movement started from the origin and ended at the origin

Total displacement is zero meters.

14. The speed between 10s and 15s is zero, because he did not move.

15. V₃ = S₃/t₃ = 100/25 = 4 m/s

16. V₁ = S₁/t₁ = 60/10 = 6 m/s   and V₄ = S₄/t₄ = 40/15 = 2.67 m/s

V₁ > V₄

God is with you!!!

4 0
3 years ago
According to the big bang theory, the universe is continually getting
oee [108]
The answer on Edge would be (A.)= Larger and Cooler ! I'm doing the same thing as y'all. Good luck everyone.

7 0
3 years ago
Read 2 more answers
A truck starts off 151 miles directly north from the city of Hartville. It travels due east at a speed of 41 miles per hour. Aft
Bess [88]

Answer:

9.51

Explanation:

The distance s is given by:

s(t) = \sqrt{151^2 + (vt)^2}

The change in distance is given by the time derivative of s:

\frac{ds}{dt} = \frac{v^2t}{\sqrt{151^2 + (vt)^2}}

For the time t you solve the equation of distance x for time:

x = vt => t = \frac{x}{v}

Plugging in for t:

\frac{ds}{dt}(t=\frac{x}{v})=\frac{vx }{\sqrt{151^2 + x^2}}=9.51

6 0
3 years ago
A bedroom bureau with a mass of 45 kg, including drawers and clothing, rests on the floor. (a) If the coefficient of static fric
sergij07 [2.7K]

Answer: the minimal force that you need to apply to move the bureau is F = 198.45N

Explanation:

If you want to move an object, you need to apply a force that is bigger than the force of the statical friction.

The force of statical friction can be written as.

Ff = k*N

where k is the coefficient of static friction, in this case, k = 0.45, and N is the normal force between the object and the surface.

In this case, the normal force is the weight of the bedroom bureau, this is:

N = m*g = 45kg*9.8m/s^2 = 441N

Then the force is:

Fr = 0.45*441N = 198.45N

This means that the minimal force that you need to apply to move the bureau is F = 198.45N

and after this point, the force of friction will work wit the kinetic coefficient of friction, that usually is smaller than the statical one.

4 0
4 years ago
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