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igomit [66]
3 years ago
6

momentum A proton interacts electrically with a neutral HCl molecule located at the origin. At a certain time t, the proton’s po

sition is h1.6 × 10−9 , 0, 0i m and the proton’s velocity is h3200, 800, 0i m/s. The force exerted on the proton by the HCl molecule is h−1.12 × 10−11 , 0, 0i N. At a time t + (2 × 10−14 s), what is the approximate velocity of the proton? answer
Physics
1 answer:
arlik [135]3 years ago
6 0

Answer:

\ m/s

Explanation:

F = Force =

m = Mass of proton = 1.7\times 10^{-27\ kg

t = Time taken = 2\times 10^{-14}\ s

Acceleration is given by

a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{}{1.7\times 10^{-27}}\\\Rightarrow a=\ m/s^2

v=u+at\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\times 2\times 10^{-14}\\\Rightarrow v=+\\\Rightarrow v=\ m/s

The velocity of the proton is \ m/s

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2 years ago
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In some areas, farmers have installed windmills to generate electricity to run the farm. Some are generating a surplus of electr
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3 years ago
Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
11111nata11111 [884]

Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

Mass of System of after collision = 1000 + 2000 = 3000 kg

Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

4 0
3 years ago
an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi
algol13

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

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El Niño occurs every 2 to 12 years. true false
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True
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