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Citrus2011 [14]
4 years ago
11

An electric field exerts an electrostatic force of magnitude 1.5 x 10-14 newton on an electron within the field. what is the mag

nitude of the electric field strength at the location of the electron?
a.2.4 x 10-33 n/c
b.1.1 x 10-5 n/c
c.9.4 x 104 n/c
d.1.6 x 1016 n/c
Physics
1 answer:
stira [4]4 years ago
7 0
The electrostatic force on the electron is given by
F=qE
where q is the electron charge and E the intensity of the electric field.

So, in order to find E, we just need to re-arrange the formula and use F=1.5 \cdot 10^{-14} N:
E= \frac{F}{q} = \frac{1.5 \cdot 10^{-14}N}{1.6 \cdot 10^{-19}C}=9.4 \cdot 10^4 N/C

So, the correct answer is C.
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Write the formula of Lever, Pulleys, wheel and axle and inclined plane.<br>​
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Answer:

Lever => d_{e} = d_{r}

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The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be
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It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>
  • It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
  • Mathematically, T²∝a³
<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>
  • The time period of geosynchronous orbit is 24 hours or 1440 minutes.
  • As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
  • If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
  • a1= (T1/T2)⅔×a2

           = (1440/90)⅔×6780

           = 43,090 km

  • Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

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Newton’s law of universal gravitation a. is equivalent to Kepler’s first law of planetary motion. b. can be used to derive Keple
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Newton published his law of universal gravitation almost a hundred
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any serious sophomore in an engineering college should be able to do,
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This demonstration is a tremendous boost for the work of both Kepler
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5 0
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The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le
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The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = \sqrt{2gh}

 V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

  = 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

     = 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

    = - 0.329 J

Hence, kinetic energy is not conserved.

8 0
3 years ago
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