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Elina [12.6K]
3 years ago
9

The periodic wave in the diagram below has a frequency of 80. hertz.

Physics
1 answer:
ASHA 777 [7]3 years ago
6 0

Answer:

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2) 5/8

4) 5/8

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A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5
svlad2 [7]
The magnetic force experienced by the proton is given by
F=qvB \sin \theta
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and \theta the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so \sin \theta=1 and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
F=ma

So we have
ma=qvB
from which we can find the magnitude of the field:
B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T
4 0
3 years ago
1. A runner has an initial speed of 2 [m/s] and slowly speeds up with a constant acceleration of
zvonat [6]

Answer:

Final Velocity = 4.9 m/s

Explanation:

We are given;. Initial velocity; u = 2 m/s

Constant Acceleration; a = 0.1 m/s²

Distance; s = 100 m

To find the final velocity(v), we will use one of Newton's equations of motion;

v² = u² + 2as

Plugging in the relevant values to give;

v² = 2² + 2(0.1 × 100)

v² = 4 + 20

v² = 24

v = √24

v = 4.9 m/s

5 0
3 years ago
If the second ball has a mass of 2.4kg and a constant acceleration of a⃗ 2= 2.8m/s2 j^ , what must the mass of the first ball be
kompoz [17]
Hope this helps you!

8 0
3 years ago
A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti
Papessa [141]

Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

Therefore, the kinetic energy lost due to friction is 22.5 J

7 0
3 years ago
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A non-_____ rock has interlocking grains with no specific pattern.
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A non <span>foliated </span>rock has interlocking grains with no specific pattern.
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4 years ago
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