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lozanna [386]
2 years ago
15

A student holds an egg outside of an open window and let's go. The window is 40 meters above the ground and the egg is falling u

nder the acceleration of gravity, which is 9.8 m/s2. There are two equations that can be used to describe its motion over time:
How far will the egg travel after 2 seconds? Be sure to use the 5 steps to solve the problem. Show which equation you used, what each variable equals, and answer the question in a complete sentence.
Physics
1 answer:
melisa1 [442]2 years ago
5 0

The distance traveled by the egg after 2 seconds is 19.6 m.

The given parameters:

  • <em>Height of the window, h = 40 m</em>

The distance traveled by the egg after 2 seconds is calculated as follows;

h = v_0_y t + \frac{1}{2} gt^2\\\\

where;

  • <em />v_0_y<em> is the initial vertical velocity of the egg = 0</em>
  • <em>g </em><em>is acceleration due to gravity = 9.8 m/s²</em>
  • <em>t </em><em>is the time of motion, = 2 s</em>

h = 0 \ + \ \frac{1}{2} \times 9.8 \times (2)^2\\\\h = 19.6 \ m

Thus, the distance traveled by the egg after 2 seconds is 19.6 m.

Learn more about second equation of motion here: brainly.com/question/7977693

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The rotating loop in an AC generator is a square 10.0 cm on each side. It is rotated at 60.0 Hz in a uniform field of 0.800 TO.
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Answer:

Explanation:

Given a square side loop of length 10cm

L=10cm=0.1m

Then, Area=L²

Area=0.1²

Area=0.01m²

Given that, frequency=60Hz

And magnetic field B=0.8T

a. Flux Φ

Flux is given as

Φ=BA Sin(wt)

w=2πf

Φ=BA Sin(2πft)

Φ=0.8×0.01 Sin(2×π×60t)

Φ=0.008Sin(120πt) Weber

b. EMF in loop

Emf is given as

EMF= -N dΦ/dt

Where N is number of turns

Φ=0.008Sin(120πt)

dΦ/dt= 0.008×120Cos(120πt)

dΦ/dt= 0.96Cos(120πt)

Emf=-NdΦ/dt

Emf=-0.96NCos(120πt). Volts

c. Current induced for a resistance of 1ohms

From ohms law, V=iR

Therefore, Emf=iR

i=EMF/R

i=-0.96NCos(120πt) / 1

i=-0.96NCos(120πt) Ampere

d. Power delivered to the loop

Power is given as

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P=-0.96NCos(120πt)•-0.96NCos(120πt)

P=0.92N²Cos²(120πt) Watt

e. Torque

Torque is given as

τ=iL²B

τ=-0.96NCos(120πt)•0.1²×0.8

τ=-0.00768NCos(120πt) Nm

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2 years ago
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