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Novay_Z [31]
3 years ago
11

A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of th

e instrument. A parallel plate capacitor sets up an electric field E which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by 3 mm and the value of the magnetic field is 0.3 T, what voltage between the plates will allow particles of speed 5 x 105 m/s to pass straight through without deflection? A. 70 V B. 140 V C. 450 V D. 1,400 V E. 2,800 V
Physics
1 answer:
Contact [7]3 years ago
5 0

Answer:

C. 450v

Explanation:

Using

Voltage= B*distance of separation*velocity

3mm x 0.3T x 5E5m/s

= 450v

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The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular re
Sever21 [200]

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

5 0
3 years ago
17. A 25 kg block is initially at rest on a rough, horizontal surface. A horizontal force of 75 N is required to set
lapo4ka [179]

Answer:

0.30581

0.24464

Explanation:

\mu_s = Coefficient of static friction

\mu_k = Coefficient of kinetic friction

F_f = 75 N

F_k = 60 N

Normal force

F_n=mg\\\Rightarrow F_n=25\times 9.81\\\Rightarrow F_n=245.25\ N

Frictional force

F_f=\mu_sF_n\\\Rightarrow \mu_s=\frac{F_f}{F_n}\\\Rightarrow \mu_s=\frac{75}{245.25}\\\Rightarrow \mu_s=0.30581

The coefficient of static friction is 0.30581

Kinetic force

F_k=\mu_kF_n\\\Rightarrow \mu_k=\frac{F_k}{F_n}\\\Rightarrow \mu_s=\frac{60}{245.25}\\\Rightarrow \mu_s=0.24464

The coefficient of kinetic friction is 0.24464

4 0
3 years ago
You have a string with a mass of 0.0133 kg. You stretch the string with a force of 8.89 N, giving it a length of 1.97 m. Then, y
Kazeer [188]

Answer:

(i) The wavelength is 0.985 m

(ii) The frequency of the wave is 36.84 Hz

Explanation:

Given;

mass of the string, m = 0.0133 kg

tensional force on the string, T = 8.89 N

length of the string, L = 1.97 m

Velocity of the wave is:

V = \sqrt{\frac{F_T}{M/L} } \\\\V = \sqrt{\frac{8.89}{0.0133/1.97} } \ = 36.29 \ m/s

(i) The wavelength:

Fourth harmonic of a string with two nodes, the wavelength is given as,

L = 2λ

λ = L/2

λ = 1.97 / 2

λ = 0.985 m

(ii) Frequency of the wave is:

v = fλ

f = v / λ

f = 36.29 / 0.985

f = 36.84 Hz

3 0
3 years ago
In designing circular rides for amusement parks, mechanical engineers must consider how small variations in certain parameters c
Bess [88]

Answer:

Part a)

dF = -\frac{mv^2}{r^2} dr

Part b)

dF = \frac{2mvdv}{r}

Part c)

dT = - \frac{2\pi r}{v^2} dv

Explanation:

Part a)

As we know that force on the passenger while moving in circle is given as

F = \frac{mv^2}{r}

now variation in force is given as

dF = -\frac{mv^2}{r^2} dr

here speed is constant

Part b)

Now if the variation in force is required such that r is constant then we will have

F = \frac{mv^2}{r}

so we have

dF = \frac{2mvdv}{r}

Part c)

As we know that time period of the circular motion is given as

T = \frac{2\pi r}{v}

so here if radius is constant then variation in time period is given as

dT = - \frac{2\pi r}{v^2} dv

8 0
3 years ago
Propose an experiment that could be used to demonstrate that friction is a contact force?
vesna_86 [32]
Maybe push or pull an object with a large amount of mass? you are force a (pushing through object) aka making contact. i hope i helped not good with physics :)
3 0
2 years ago
Read 2 more answers
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