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ZanzabumX [31]
3 years ago
5

List the 5 Greek variables used to measure rotational motion and their counterparts for translational motion

Physics
1 answer:
Leni [432]3 years ago
4 0

Explanation:

pizs cake and they dont

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You find yourself in the middle of a frozen lake with a surface so slippery (ms = mk = 0) you cannot walk. However, you happen t
EleoNora [17]

Answer:Maybe if you would slide on your stomach you can get out of there.

Explanation:

8 0
4 years ago
A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume th
Katyanochek1 [597]

Answer:

9.88 milivolt

Explanation:

Given: diameter d = 5.2 cm

magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T

t = 0.29 sec.

we know emf =  - dΦ/dt

and flux Φ = BA

A= area

therefore emf ε = -A(B_2-B_1)/Δt

=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV

3 0
3 years ago
Explain how a hydroelectric power plant converts energy stored in water into electricity
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Hot water, as in very hot water, goes to steam with an enormous change in volume and pressure. This is capable of driving turbine blades, which in turn rotate enormous copper (eg) wired coils in enormous magnetic fields. This in turn produces electricity via electromagnetic induction and Fleming's right hand (I think it is) DYNAMO rule. That goes down miles and mile of v heavy conducting wire/cable as electricity.
4 0
4 years ago
For a short time the missile moves along the parabolic path y=(18−2x2) km. If motion along the ground is measured as x=(4t−3) km
muminat

Answer with Explanation:

We are given that

y=(18-2x^2) km

x=(4t-3)km

Differentiate x and y w.r.t t

\frac{dx}{dt}=4

\frac{dy}{dt}=-4x\frac{dx}{dt}

\frac{dy}{dt}=-4x\times 4=-16x=-16(4t-3)

v_x=\frac{dx}{dt}=4

v_y=\frac{dy}{dt}=-16(4t-3)

Substitute t=1

v_x=4

v_y=-16(4-3)=-16

Magnitude of velocity=\mid v\mid=\sqrt{v^2_x+v^2_y}

\mid v\mid=\sqrt{4^2+(-16)^2}=16.49 m/s

Hence, the magnitude of the missile's velocity=16.49 m/s

a_x=\frac{d(\frac{dx}{dt})}{dt}=\frac{d(4)}{dt}=0

a_y=\frac{d(\frac{dy}{dt})}{dt}=\frac{d(-16(4t-3))}{dt}=-64

Substitute t=1

a_x=0,a_y=-64

\mid a\mid=\sqrt{a^2_x+a^2_y}

\mid a\mid=\sqrt{0+(-64)^2}=64m/s^2

Hence, the magnitude of acceleration when t=1 s=64m/s^2

3 0
3 years ago
An atom emission of light with a specific amount of energy confirms that ?
marusya05 [52]

The correct answer is B, electrons emit and absorb energy based on their position around the nucleus.

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3 years ago
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