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____ [38]
3 years ago
7

A proton is placed in an electric field of intensity 700 N/C. What are the magnitude and direction of the acceleration of this p

roton due to this field? (mproton = 1.67 × 10-27 kg, e = 1.60 × 10-19 C)
Physics
1 answer:
olya-2409 [2.1K]3 years ago
6 0

Answer:

Acceleration of proton will be a=0.67\times 10^{11}m/sec^2

Explanation:

We have given a proton is placed in an electric field of intensity of 700 N/C

So electric field E = 700 N/C

Mass of proton m=1.67\times 10^{-27}kg

Charge on proton e=1.6\times 10^{-19}C

So electric force on the proton F=qE=1.6\times 10^{-19}\times 700=1.120\times 10^{-16}N

This force will be equal to force due to acceleration of the proton

According to newton's law force is given by F = ma

So 1.67\times 10^{-27}\times a=1.120\times 10^{-16}

a=0.67\times 10^{11}m/sec^2

So acceleration of proton will be a=0.67\times 10^{11}m/sec^2

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Answer:

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Explanation:

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The synthesis of nitrogen trihydride from nitrogen gas and hydrogen gas is shown by which balanced chemical equation?
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A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec
Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 8=50.24rad/sec

Velocity of the string wave is equal to v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec

Power of wave propagation is equal to P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt

So power of the wave will be equal to 5.464 watt

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3 years ago
A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with
Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = \dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2

              = \dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2

              = 31 J

Final KE = \dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J

Loss in KE = 31 J - 25.6 J = 5.4 J.

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