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Yuliya22 [10]
2 years ago
10

A gardener uses 2700 J of work to push a wheelbarrow for 60 seconds. How much power does she use?

Physics
2 answers:
Phantasy [73]2 years ago
5 0

Answer:

I think 90 w I am not sure but I think it is right

I hope it is helpful for u

Zina [86]2 years ago
3 0

Answer:

<h2><em>4</em><em>5</em><em> </em><em>W</em></h2>

<em>Sol</em><em>ution</em><em>,</em>

<em>Work</em><em>(</em><em>W</em><em>)</em><em>=</em><em>2</em><em>7</em><em>0</em><em>0</em><em> </em><em>Joule</em>

<em>Time</em><em>(</em><em>t</em><em>)</em><em>=</em><em>6</em><em>0</em><em> </em><em>Seconds</em>

<em>Power</em><em>=</em><em>?</em>

<em>Now</em><em>,</em>

<em>power =  \frac{work \: done}{time \: taken}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{2700}{60}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 45 \: \: watt</em>

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em>

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A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn
Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

F=mrw^2

m= mass of the corresponding bugs

r= corresponding radial distance of each bug

w= angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

F ∝ mr

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0
2 years ago
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Which muscle disorder does donna’s doctor diagnose? donna is suffering from leg pain in which the calf muscles contract involunt
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Donna's doctor must have diagnosed her with muscle cramps. With the given symptoms above, it is likely that she is experiencing muscle cramps. Muscle cramps most likely happen when the individual experiencing it is lack of fluid intake, causing the muscles to tighten causing pain and for the muscle to contract involuntarily.
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2 years ago
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You throw a baseball straight up into the air with a speed of 24.5 m/s. How long does it take the baseball to reach its highest
galben [10]
Time=speed/acceleration
Gravitaional Acceleration=9.8 m/s^2
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3 0
2 years ago
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The blue curve is the plot of the data. The straight orange line is tangent to the blue curve at t = 40 s. A plot has the concen
Sever21 [200]

Answer:

  0.00325 moles/liter/second

Explanation:

The tangent line has a slope of (y2 -y1)/(x2 -x1) = (0.35-0.48)/(40-0) = -0.00325.

The rate of the reaction is about 0.00325 moles/liter/second.

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This is the rate of decrease of the concentration of A.

5 0
2 years ago
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A worker stands still on a roof sloped at an angle of 35° above the horizontal. He is prevented from slipping by static friction
aleksley [76]

Answer:

99.63 kg

Explanation:

From the force diagram

N = normal force on the worker from the surface of the roof

f = static frictional force = 560 N

θ = angle of the slope = 35

m = mass of the worker

W = weight of the worker = mg

W Cosθ = Component of the weight of worker perpendicular to the surface of roof

W Sinθ = Component of the weight of worker parallel to the surface of roof

From the force diagram, for the worker not to slip, force equation must be

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m = 99.63 kg

5 0
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