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Rzqust [24]
3 years ago
7

100 POINTS!!!!!!

Physics
1 answer:
Paha777 [63]3 years ago
5 0

Answer:

How do you stop answe question?

Explanation:

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The efficiency of a motor is 15%. The student calculated the useful output energy transfer is 1.20J. Calculate the total input e
Margaret [11]

Answer:

8.0\; \rm J.

Explanation:

The efficiency of a machine is the ratio between the useful output and the energy input:

\begin{aligned}\text{efficiency} &= \frac{\text{useful output}}{\text{energy input}} \times 100\%\end{aligned}

Rearrange this equation to find energy input in terms of efficiency and useful output:

\displaystyle \text{energy input} = \frac{\text{useful output}}{\text{efficiency} / (100\%)}.

Substitute in the values: \text{useful output} = 1.2\; \rm J and \text{efficiency} = 15\%. Evaluate to find the value of \text{energy input}:

\begin{aligned} \text{energy input} &= \frac{\text{useful output}}{\text{efficiency} / (100\%)} \\ &= \frac{1.20\; \rm J}{15\% / (100\%)} \\ &= 8.0\; \rm J\end{aligned}.

(Rounded to two significant figures as in the value of efficiency.)

4 0
3 years ago
A 15 m long garden hose han an inner diameter of 2.5 cm. One end is connected to a spigot; 20 C water flows from the other end a
pogonyaev
We can solve this problem using <span>Hagen–Poiseuille equation. Derivation of this equation is a bit complicated so I will just write down the equation.
</span>\Delta P= \frac{8\mu QL}{\pi R^4}
This equation gives you the pressure drop <span>in an </span>incompressible<span> and </span>Newtonian<span> fluid in </span>laminar flow<span> flowing through a long cylindrical pipe of the constant cross section.
L is the length of the cylinder, Q is the volumetric flow rate, R is the radius of the pipe, and \mu is dynamic viscosity.
Dynamic viscosity of water at 20 Celsius is 0.001 PaS. 
Now we can calculate the pressure drop:
</span>\Delta P= \frac{8\cdot 0.001\cdot 15 \cdot 0.015}{\pi 3.9\cdot 10^{-7}}=1469.12 $Pa<span>
</span>
7 0
3 years ago
What happens as the moon moves through earth's shadow
matrenka [14]
Pushing pool tides something about the ocean
7 0
3 years ago
There are elements in the surface of the sun that do not exist on earth. true false
suter [353]
I think this is false
7 0
3 years ago
Two objects are dropped from rest from the same height. Object A falls through a distance during a time t, and object B falls th
UNO [17]

Answer:

Distance covered by B is 4 times distance covered by A

Explanation:

For an object in free fall starting from rest, the distance covered by the object in a time t is

s=\frac{1}{2}gt^2

where

s is the distance covered

g is the acceleration due to gravity

t is the time elapsed

In this problem:

- Object A falls through a distance s_A during a time t, so the distance covered by object A is

s_A=\frac{1}{2}gt^2

- Object B falls through a distance s_B during a time 2t, so the distance covered by object B is

s_B=\frac{1}{2}g(2t)^2 = 4(\frac{1}{2}gt^2)=4s_A

So, the distance covered by object B is 4 times the distance covered by object A.

5 0
3 years ago
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