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3 years ago

What is the force of friction acting on this box? N What is the net force acting on this box? N to the

Physics

2 answers:

Zanzabum3 years ago

5 0

Answer: -3, 12, right

Explanation:

n/a just did the test

e-lub [12.9K]3 years ago

3 0

Answer:

-3,12,right

Explanation:

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Your laundry basket weighs 22 N and your room is 3.0 m above you on the second floor. It takes you 6.0 seconds to carry the laun

Fofino [41]

<h2>Power is 11 W</h2>

Explanation:

Power = Work ÷ Time

Work = Force x Displacement

Force = 22 N

Displacement = 3 m

Time = 6 seconds

Substituting

Work = Force x Displacement

Work = 22 x 3 = 66 J

Power = Work ÷ Time

Power = 66 ÷ 6

Power = 11 W

Power is 11 W

8 0

3 years ago

What is another name for kidney stones?

Naddika [18.5K]

Another (slightly more technical) name for kidney stones is "renal calculus".

I know this because we have met many times, and we are old friends.

8 0

3 years ago

Read 2 more answers

A sprinter with a mass of 80 kg accelerates from 0 m/s to 9 m/s in 3 s. What is the runner's acceleration?

sertanlavr [38]

Acceleration = (change in speed) / (time for the change) = 9/3 = <em>3 m/s²</em> .

His mass makes no difference.

5 0

3 years ago

Calculate the density of a material that has a bat a mass of 52.457 g any volume of 13.5 cm³

3241004551 [841]

Answer:
d = 3.8857 g/cm^3

Explain:
Formula/: d = m/V

Hope this helps!
(a brainliest would be appreciated)

4 0

3 years ago

We are going to make an imaginary engine using water. We are going to heat 100 grams of water to 120 C from its initial temperat

Svetach [21]

Answer:

The work done by this engine is 800 cal

Explanation:

Given:

100 g of water

120°C final temperature

22°C initial temperature

30°C is the temperature of condensed steam

Cw = specific heat of water = 1 cal/g °C

Cg = specific heat of steam = 0.48 cal/g °C

Lw = latent heat of vaporization = 540 cal/g

Question: How much work can be done using this engine, W = ?

First, you need to calculate the heat that it is necessary to change water to steam:

$Q_{1} =m_{w} C_{w} (100-22)+m_{w}L_{w}+m_{w}C_{g}(120-100)$

Here, mw is the mass of water

$Q_{1} =(100*1*78)+(100*540)+(100*0.48*20)=62760cal$

Now, you need to calculate the heat released by the steam:

$Q_{2} =m_{w}C_{g}(120-100)+m_{w}L_{w}+m_{w}C_{w}(100-30)=(100*0.48*20)+(100*540)+(100*1*70)=61960cal$

The work done by this engine is the difference between both heats:

$W=Q_{1}-Q_{2}=62760-61960=800cal$

8 0

3 years ago