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vichka [17]
3 years ago
10

Ok so how do i determine how many point i wanna give out on a question

Physics
2 answers:
Leviafan [203]3 years ago
7 0

Explanation:

when ever you ask a question and you pick the subject and the grade level it shows points you will see 10 and above numbers so that is how know how many points you given have a good day

Sergeu [11.5K]3 years ago
4 0
Sometimes it just picks for you
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paper by itself catches fire easily whereas a piece of paper wrapped around an alluminium pipe does not.​
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The paper does not catch fire when wrapped around aluminium pipe because aluminium absorbs the heat, so paper does not attain its ignition temperature.

Explanation:

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True or false: in applying the pythagorean theorem to vector addition, the resultant vector is given by the perpendicular vector
olganol [36]

It is False in applying the pythagorean theorem to vector addition, the resultant vector is given by the perpendicular vector

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1 year ago
a vehicle changes its velocity from 90 km/h to a dead stop in 10s. show that its acceleration in doing so is -2.5m/s^2
telo118 [61]

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4 0
4 years ago
An electron of kinetic energy 1.59 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 35.4 cm
Blizzard [7]

Answer:

a) v = 2.36 \cdot 10^{7} m/s

b) B = 3.80 \cdot 10^{-4} T

c) f = 1.06 \cdot 10^{7} Hz

d) T = 9.43 \cdot 10^{-8} s

Explanation:

a) We can find the electron's speed by knowing the kinetic energy:

K = \frac{1}{2}mv^{2}

Where:    

K: is the kinetic energy = 1.59 keV

m: is the electron's mass = 9.11x10⁻³¹ kg

v: is the speed =?

v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2*1.59 \cdot 10^{3} eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}}{9.11 \cdot 10^{-31} kg}} = 2.36 \cdot 10^{7} m/s

b) The electron's speed can be found by using Lorentz's equation:

F = q(v\times B) = qvBsin(\theta)   (1)

Where:

F: is the magnetic force

q: is the electron's charge = 1.6x10⁻¹⁹ C

θ: is the angle between the speed of the electron and the magnetic field = 90°

The magnetic force is also equal to:

F = ma_{c} = m\frac{v^{2}}{r}   (2)

By equating equation (2) with (1) and by solving for B, we have:

B = \frac{mv}{rq} = \frac{9.11 \cdot 10^{-31} kg*2.36 \cdot 10^{7} m/s}{0.354 m*1.6 \cdot 10^{-19} C} = 3.80 \cdot 10^{-4} T

c) The circling frequency is:

f = \frac{1}{T} = \frac{\omega}{2\pi} = \frac{v}{2\pi r}

Where:

T: is the period = 2π/ω

ω: is the angular speed = v/r

f = \frac{v}{2\pi r} = \frac{2.36 \cdot 10^{7} m/s}{2\pi*0.354 m} = 1.06 \cdot 10^{7} Hz

d) The period of the motion is:

T = \frac{1}{f} = \frac{1}{1.06 \cdot 10^{7} Hz} = 9.43 \cdot 10^{-8} s

I hope it helps you!

8 0
3 years ago
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