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zhannawk [14.2K]
3 years ago
10

Bill and Nageen have each built an electric motor and they are testing them by having them lift boxes. Bill's motor lifts a box

0.35 metres in 2 seconds. Nageen's motor lifts the same box 0.35 metres in 1.8 seconds. Whose motor has applied more power to the box?
Physics
1 answer:
photoshop1234 [79]3 years ago
5 0

Answer:

Bill's motor power: W_B = F x S / T = F x 0.35 / 2= 0.175F

Nageen's motor power: W_N = F x S / T = F x 0.35 / 1.8 = 0.194F

=> 0.194F > 0.175F => Nageen's motor applied more power to the box than Bill's motor.

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Why does Quito, Equator has very little changes to the daylight hours<br> throughout the year?
marin [14]

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An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce
Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Frequency of pendulum is given by

f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}

Given in the question

f_p=\dfrac{1}{2}f_s

\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m

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6 0
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Consider a single-slit diffraction pattern for λ=589nm , projected on a screen that is 1.00 m from a slit of width 0.25 mm. How
const2013 [10]

Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm

Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as

y = R×(mλ/d)

Where y = distance between nth fringes and the center fringe.

m = order of fringe

λ = wavelength of light = 589nm = 589×10^-9m

R = distance between slits and screen = 1.0m

d = distance between slits = 0.25mm = 0.00025m

For distance between the first dark fringe and the center fringe.

This implies that m = 1

y = 1 × 589×10^-9 × 1/0.00025

y = 589×10^-9/0.00025

y = 2,356,000 × 10^-9

y = 2.356 × 10^-3 m = 2.356mm

For the second dark fringe, this implies that m = 2

y = 1 × 2 × 589×10^-9/0.00025

y = 1178 × 10^-9 /0.00025

y = 4,712,000 × 10^-9

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8 0
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DerKrebs [107]

Answer:

The conditions necessary for hearing the echo. The distance between the sound source and the reflecting surface must not be less than 17 metres where the time period between hearing the original sound and its echo should not be less than 0.1 of a second.

6 0
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