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Novay_Z [31]
3 years ago
14

An object with a mass of 30 kg moves at a rate of 15 m/S over distance of 10 m. 900 joules of work was done in 10 seconds. What

was the power generated by this project?
Physics
1 answer:
Sliva [168]3 years ago
7 0

Answer:

P = 90 W

Explanation:

Mass of an object, m = 30 kg

Speed of the object, v = 15 m/s

Distance, d = 10 m

Work done, W = 900 J

Time, t = 10 s

We need to find the power generated by this project. We know that power is equal to work done divided by time taken.

So,

P=\dfrac{W}{t}\\\\P=\dfrac{900\ J}{10\ s}\\\\P=90\ W

So, the required power generated is 90 W.

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Sean and Tommy are debating the positive and negative impacts of technology on environmental quality. Tommy is listing the negat
Pachacha [2.7K]

Answer:

Factories contribute to air and water pollution.

Explanation:

8 0
3 years ago
A bartender slides a beer mug at 1.6 m/s towards a customer at the end of a frictionless bar that is 1.1 m tall. The customer ma
Drupady [299]

Answer:

(a): the mug hits the floor 0.752m away from the end of the bar.

(b): the speed of the mug at impact are:

V= 4.87 m/s

direction= 70.82º below the horizontal.

Explanation:

Vx= 1.6 m/s

Vy=?

h= 1.1 m

g= 9.8 m/s²

t is the fall time

t=\sqrt{\frac{2*h}{g} }

t=0.47 sec

Vy= g*t

Vy= 4.6 m/s

V=\sqrt{Vx^{2} +Vy^{2}

V= 4.87 m/s

α= tan⁻¹(Vy/Vx)

α= -70.82º

6 0
3 years ago
Read 2 more answers
How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un
Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is 4.20\times10^{8}\ kg-m/s.

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

W=K.E

W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2

Put the value into the formula

W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2

W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2

W=1.122\times10^{-9}\ J

W=7001\ MeV

(b). We need to calculate  the momentum of this proton

Using formula of momentum

p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}

Put the value into the formula

p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}

p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}

p=1.404\times10^{-26}c

p=4.20\times10^{8}\ kg-m/s

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is 4.20\times10^{8}\ kg-m/s.

4 0
3 years ago
How does a rubber rod become negatively charged through friction?
Artist 52 [7]
Thank you for posting your question here and Brainly!~

Even though you have not provided answer choices, I believe the answer is whatever letter corresponds with the answer: "It is rubbed with another object, and electrons move onto the rod."

Hope I helped!~ Brainliest appreciated.
8 0
4 years ago
Read 2 more answers
A vector is given in terms of its magnitude, v=8.7m/s and its direction, 256 degrees standard position. a) what is this vector i
Wittaler [7]

Answer:

Explanation:

Given

magnitude of vector=8.7 m/s

Direction =256^{\circ}

Therefore vector x component is =8.7\cos (256)

=-2.104

i.e. it is acting in negative x axis

vector y component=8.7\sin (256)=-8.44 m/s

Therefore it lies in negative y axis

Judging vector x and y component

It lies in third quadrant

4 0
3 years ago
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