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Naily [24]
3 years ago
7

An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma

gnetic field with a magnitude of 3.70×10−3 t .
Physics
1 answer:
Leya [2.2K]3 years ago
4 0
Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
F=qvB \sin \theta
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
\theta the angle between the directions of v and B.

Re-arranging the formula, we find:
v= \frac{F}{qB \sin \theta}
and by substituting the data of the problem (the charge of the electron is q=1.6 \cdot 10^{-19} C), we find the velocity of the electron:
v= \frac{F}{qB \sin \theta}= \frac{4.90 \cdot 10^{-15}N}{(1.6 \cdot 10^{-19}C)(3.70 \cdot 10^{-3} T)(\sin 60^{\circ})}=9.56 \cdot 10^6 m/s
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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and
Svetlanka [38]

Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

u = Initial velocity = 4 m/s

s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

S= 1.632-4.9(0.166)

S = 1.632-0.815

S = 0.817m

Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

0.554 =t^2

t =√0.554

t = 0.744s

Hence, her feet were in the air for 0.744+0.408seconds

= 1.152s

Also recall from equation 1

V= u+at

V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

4 0
2 years ago
QUESTION:
vampirchik [111]

Answer:

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5 0
2 years ago
A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec
Phoenix [80]

Answer:

The particle path will follow

(d) a circular path

Explanation:

When a charged particle having charge of magnitude 'q' enters into a magnetic field such that its velocity vector '\vec{V}' is perpendicular to the direction of the magnetic field '\vec{B}', then it will experience a force, called Lorentz force (\vec{F_{L}}), given by

\vec{F_{L}} = \vec{V} \times \vec{B}

As shown in the figure, the magnetic field is directed perpendicular to the plane and towards the plane (as shown by the circle and 'X'-sign) and the velocity vector is from left to right on the plane.

According to the property of cross-product, the Lorentz force (F_{L}) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path,as shown in the figure.

7 0
3 years ago
What is Potential Energy
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Answer:

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Explanation:

8 0
3 years ago
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A runner of mass 56.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its cent
SCORPION-xisa [38]

Answer: -0.84 rad/sec (clockwise)

Explanation:

Assuming no external torques act on the system (man + turntable), total angular momentum must be conserved:

L1 = L2

L1 = It ω + mm. v . r = 81.0 kg . m2 .21 rad/s – 56.0 kg. 3.1m/s . 3.1 m  

L1 = -521.15 kg.m2/sec (1)

(Considering to the man as a particle that is moving opposite to the rotation of  the turntable, so the sign is negative).

Once at rest, the runner is only a point mass with a given rotational inertia respect from the axis of rotation, that can be expressed as follows:

Im = m. r2 = 56.0 kg. (3.1m)2 = 538.16 kg.m2

The total angular momentum, once the runner has come to an stop, can be written as follows:

L2= (It + Im) ωf = -521.15 kg.m2/sec  

L2= (81.0 kg.m2 + 538.16 kg.m2) ωf = -521.15 kg.m2/sec  

Solving for ωf, we get:

ωf = -0.84 rad/sec  (clockwise)

5 0
3 years ago
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