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Arturiano [62]
3 years ago
8

How much money does a mixer use if the efficiency is 50% and it wastes 20J of energy ?

Physics
1 answer:
GalinKa [24]3 years ago
5 0

Answer:

Ei = 40J

Explanation:

Given the following data;

Efficiency = 50%

Energy output, Eo = 20 J

To find the amount of energy used by the mixer:

Efficiency = Eo/Ei * 100

Where;

Eo is the output energy.

Ei is the input energy.

Substituting into the formula, we have;

50 = 20/Ei * 100

Cross-multiplying, we have;

50Ei = 2000

Ei = 2000/50

Ei = 40J

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Using the strap at an angle of 31.0° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15.0
lora16 [44]

Answer:

<h2>154.73N</h2>

Explanation:

The question is incomplete. Here is the complete question.

Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.

Check the diagram related to the question in the attachment below for better understanding.

The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.

The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).

Ty = 15sin31°

Ty = 7.73N

W = mass * acceleration due to gravity

W = 15.0*9.8

W = 147N

The normal force is therefore expressed as;

N = Ty + W

N = 7.73 + 147

N = 154.73N

3 0
3 years ago
The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of
kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

\rho_{0\°C}= 730 kg/m^{3} is the density of gasoline at 0\°C

\beta=9.60(10)^{-4} \°C^{-1} is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to m^{3}:

V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3} (1)

Knowing density is given by: \rho=\frac{m}{V}, we can find the mass m_{1} of 8.50 gallons:

m_{1}=\rho_{0\°C}V

m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg (2)

Now, we have to calculate the factor f by which the volume of gasoline is increased with the temperature, which is given by:

f=(1+\beta(T_{f}-T_{o})) (3)

Where T_{o}=0\°C is the initial temperature and T_{f}=21.7\°C is the final temperature.

f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C)) (4)

f=1.020832 (5)

With this, we can calculate the density of gasoline at 21.7\°C:

\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)

\rho_{21.7\°C}=745.207 kg/m^{3} (6)

Now we can calculate the mass of gasoline at this temperature:

m_{2}=\rho_{21.7\°C}V (7)

m_{2}=(745.207 kg/m^{3})(0.0323 m^{3}) (8)

m_{2}=24.070 kg (9)

And finally calculate the mass difference \Delta m:

\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg (10)

\Delta m=0.4911 kg (11) This is the extra mass of gasoline

6 0
3 years ago
A satellite of 2,140 kg travels around the earth, of 5.97 - 1024
77julia77 [94]

Answer:

8,961

Explanation:

3 0
3 years ago
Need help! Need help! Need help! Need help! Need help! Need help!
Maslowich

Answer:

i can help you i know this answer

5 0
3 years ago
Why is an element considered a pure substance?
anastassius [24]

Answer:

An element is a pure substance and is made of only one type of atom.

Explanation:

It cannot be broken down into a simpler substance.

3 0
2 years ago
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