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Effectus [21]
4 years ago
11

Which device is using a motor? A. A water heater creates heat energy from electric energy. B. A waterfall rotates a waterwheel t

o turn an axle. C. A body massager uses electric energy to move back and forth. D. A solar charger creates electric energy from the Sun.
Physics
1 answer:
Mumz [18]4 years ago
4 0

Answer:

C is the right answer.

Body massager uses electrical energy to move back and forth. In this sense, a motor is being used for the operation

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Answer:

Double Replacement.

Explanation:

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What does your ecological foot print tell you?
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How big your foot is. (Big Brain)

Explanation:

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Because communication involves two or more people acting in both sender and receiver roles and because their messages are depend
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Transactional process

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3 years ago
A businesswoman is rushing out of a hotel through a revolving door with a force of 80 N applied at the edge of the 3 m wide door
mel-nik [20]

Answer:

Explanation:

Given

Force applied F=80\ N

Door is d=3\ m wide

for gate to revolve Properly Pivot Point must be at center i.e. 1.5 from either end

Torque applied is T=force\times distance

Maximum torque

T_{max}=F\times \frac{d}{2}

T_{max}=80\times \frac{3}{2}

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7 0
3 years ago
A 16.5-kg crate starts at rest at the top of a 60.0° incline. The coefficients of friction are μs = 0.400 and μk = 0.300. The cr
irga5000 [103]

Answer:

t = 1.62 s

Explanation:

given,

mass of the block m₁ = 16.5 Kg

m₂ = 8 Kg

angle of inclination = 60°

μs = 0.400 and μk = 0.300

time to slide 2 m = ?

a) let a is the acceleration of the block m₁ downward.

Net force acting on m₂,

F₂ = T - m₂ g

m₂a = T - m₂ g

a = \dfrac{T}{m_2} - g.......(1)

net force acting on m₁

F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T

m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T

a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}.........(2)

from equations 1 and 2

\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}

\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}

 T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}  

T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}

T = 90.61 N

from equation (1)

a = \dfrac{90.61}{8} - 9.8.......(1)

a = 1.52 m/s²

let t is the time taken

Apply,

d = ut + 0.5 a t²

2 = 0 + 0.5 x 1.52 x t²

t = \sqrt{2.63}

t = 1.62 s

5 0
3 years ago
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