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irina [24]
3 years ago
7

How long would it take a machine to do 5000 J of work if the power rating of the machine is 100 watts?

Physics
1 answer:
NeX [460]3 years ago
4 0

Answer:

25 N; 250 W.

Explanation: Work Done

=

Force

×

Displacement

×

cos

(

The angle between Force and Displacement

)

So, Let's Assume the Force to be

x

Newtons.

So, According to The Sum,

×

x

x

⋅

100

=

2500

⇒

x

=

25

00

1

00

=

25

So, The force was

25

N

.

And, We also know,

Power

=

Work Done

Time

So,

The Power of the Machine =

2500

10

Watts

=

250

Watts

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When is a secondary source more helpful than a primary source?
UNO [17]

Answer:

I think the answer is C.

Explanation:

A primary source is a first hand account of an event while a secondary source is a retelling or second hand account meaning as many details will be prevalent.

8 0
2 years ago
A table tennis ball with a mass of 0.003 kg and a soccer ball with a mass of 0.43 kg or both Serta name motion at 16 M/S calcula
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8 0
3 years ago
Read 2 more answers
Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show
Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

m_2 = \frac{20}{3} kg

Part b)

Angular acceleration is given as

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = 176.6 N

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

m_1g r_1 = m_2 g r_2

20 g(0.5) = m_2 g(1.5)

10 = 1.5 m_2

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Part b)

When block m_2 is removed then system becomes unstable

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m_1g - T = m_1 a

also we have

T r_1 = I\alpha

now we have

m_1g = \frac{I a}{r_1^2} + m_1 a

a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}

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\alpha = \frac{a}{r_1}

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\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

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7 0
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