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2 years ago

6

A homeowner wants to reduce the environmental damage of the energy source used to heat her home.

2 answers:

koban [17]2 years ago

5 0

Answer:

WIND

Explanation:

EDG 2021

liberstina [14]2 years ago

4 0

Answer:

wind

Explanation:

wind is the only one of those onto that list that is a renewable source

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Suppose you exert 150 n on your refrigerator and push it across the kitchen floor at constant velocity. what friction force acts

Pachacha [2.7K]

Due that the velocity is constant that means that friction force is equal to the force exert by you, otherwise the refrigerator will accelerate or decelerate and in both cases velocity will not be constant.
So then the friction force between refrigerator and floor is 150 Newtons.

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3 years ago

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Urgent! Which is not an example of work? Question 2 options: pushing a box across the floor picking up a box off the floor raisi

OLga [1]

Answer:

trying to push a rock that never moves

Explanation:

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2 years ago

bagirrra123 [75]

Explanation:

The reading on the scale is

W = m(g + a)

= (77 kg)(9.8 m/s^2 + 2 m/s^2)

= 908.6 N

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2 years ago

An object is represented by the dot on the motion map. Which object is most likely represented by the dot?

crimeas [40]

B) I just took the test and put D but it gave me the wrong answer. It told me it was B.

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2 years ago

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Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

2 years ago