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3 years ago

What would an automobile engineer MOST LIKELY suggest to improve the efficiency of a car’s engine?

Physics

1 answer:

n200080 [17]3 years ago

8 0

The answer is only yes cause they try to make it run more efficient

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A pump of power 1500W is used for 5 mins the energy supplied by the pump is

Anika [276]

Answer:

300W

Explanation:

from my years in middle school, i have learned the equation is energy =power*time is 300W

7 0

4 years ago

A student sees a newspaper ad for an apartment that has 1330 square feet of floor space. How many square meters of area are ther

hichkok12 [17]

One square meter equals 10.7 feet.

$1330 ft^2 * \frac{1m}{10.7ft^2} = 123.561$

8 0

4 years ago

Questions on the photo

White raven [17]

Answer:

Option C: Third Class

Explanation:

This is third class because the effort or the input force is in the middle between the fulcrum and the load.

8 0

3 years ago

A Decay Chain

Strike441 [17]

Answer : The energy released in first step of thorium-232 decay chain is $7.974\times 10^{-13}J$

Explanation :

First we have to calculate the mass defect $(\Delta m)$ .

The balanced reaction is,

$^{232}Th\rightarrow ^{228}Ra+^{4}He$

Mass defect = Sum of mass of product - sum of mass of reactants

$\Delta m=(\text{Mass of Ra}+\text{Mass of He})-(\text{Mass of Th})$

$\Delta m=(228.0301069+4.002602)-(232.038054)=5.34\times 10^{-3}amu=8.86\times 10^{-30}kg$

conversion used : $(1amu=1.66\times 10^{-27}kg)$

Now we have to calculate the energy released.

$Energy=\Delta m\times (c)^2$

$Energy=(8.86\times 10^{-30}kg)\times (3\times 10^8m/s)^2$

$Energy=7.974\times 10^{-13}J$

The energy released is $7.974\times 10^{-13}J$

Therefore, the energy released in first step of thorium-232 decay chain is $7.974\times 10^{-13}J$

7 0

3 years ago

You are asked to help design a new type of loop-the-loop ride. Instead of rolling down a long hill to generate the speed to go a

Trava [24]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

5 0

4 years ago