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yanalaym [24]
3 years ago
10

Which of the following statements about the force on a charged particle due to a magnetic field are not valid?

Physics
1 answer:
Vinil7 [7]3 years ago
5 0
The correct answer is "None of the above; all of these statements are valid." All the statements namely, it depends on the particle's charge, it depends on the strength of the external magnetic field, it depends on the particle's velocity, and it acts at right angles to the direction of the particle's motion are all valid. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help. 
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A negative slope on the velocity vs. time graph indicates that the object is not accelerating​
liubo4ka [24]

AnswerNo

However, the acceleration is the opposite of the movement:

Explanation:

8 0
3 years ago
Un libro del peso di 12 N è in equilibrio su un tavolo. Sapendo che il coefficiente di attrito statico vale 0,5, la forza di att
Tanzania [10]

Answer:

60

Explanation:

Translation -

A book weighing 12 N is balanced on a table. Knowing that the static friction coefficient is 0.5, how much is the friction force worth?

Friction force is

f = u * n

f = 0.5 * 12N

f = 60

4 0
3 years ago
An airport has runways only 198 m long. a small plane must reach a ground speed of 39 m/s before it can become airborne. what av
Anarel [89]
S: 198 m 
v=39 m/s 
u=0
t=? 
a=?

v²=u²+2as
(39)²=(0)²+2(a)(198) 
1521=396a
1521/396=a
3.84 m/s^2 = a 

Hope I helped :) 
8 0
3 years ago
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the
Zepler [3.9K]

Answer:

51.94°

Explanation:

I_0 = Unpolarized light

I_2 = Light after passing though second filter = 0.19I_0

Polarized light passing through first filter

I_1=\frac{I_0}{2}

Polarized light passing through second filter

I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}

The angle between the two filters is 51.94°

5 0
2 years ago
PLEASE HELP. I WILL MARK BRAINIEST
anyanavicka [17]
Aaaaaaaaaaaaaaaaaaaaaaaa
6 0
3 years ago
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