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a_sh-v [17]
2 years ago
12

In one hour, coal supplies 500 000 J of energy. The energy amounts to 380 000 J. How much useful energy is produced in one hour?

Physics
1 answer:
Debora [2.8K]2 years ago
5 0

Answer:

120,000J

Corrected question;

In one hour, coal supplies 500 000 J of energy. The wasted energy amounts to 380 000 J. How much useful energy is produced in one hour?

Explanation:

Given;

Total energy Et = 500,000 J

Wasted Energy Ew = 380,000J

The amount useful energy is the amount of energy that is available for supply.

This can be derived by subtracting the wasted energy from the total energy.

Useful energy = Total Energy - wasted energy

Eu = Et - Ew

Substituting the given values;

Eu = 500,000J - 380,000

Eu = 120,000 J

The amount of useful energy produced in one hour is 120,000 J

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If you run North for 5 meters and then East for 15 meters and then
kvasek [131]

Answer:

1m/s

Explanation:

D = 22m

T = 22s

S = 22÷22

= 1 m/s

7 0
3 years ago
SOHCAHTOA MATH REVIEW <br><br> Find the angle x from the diagram above
mestny [16]

Explanation:

Hey there!

Here,

It is a Right angled triangle.

Now, Taking angle x as a reference angle.

We get,

perpendicular (p) = 24.

Hypotenuse (h) = 29.

Using sin ratio we get.

\sin( \alpha )  =  \frac{p }{h}

Now, keep values and simplify them.

\sin(x)  =  \frac{24}{29}

sinx = 0.8275.

x =  { \sin }^{  - 1}  (0.8275)

Therefore the value of x is 55.842°. Or 55°.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

5 0
3 years ago
A spherical balloon has a radius of 6.95m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of ai
klio [65]

Answer:

602.27 kg

Explanation:

The computation of the largest mass of cargo the balloon can lift is shown below:-

Volume of helium inside the ballon= (4 ÷ 3) × π × r^3

= (4 ÷ 3) × 3.14 × 6.953

= 1406.19 m3

Mass the balloon can carry = volume × (density of air-density of helium)

= 1406.19 × (1.29-0.179)

= 1562.27 kg

Mass of cargo it can carry = Mass it can carry - Mass of structure

= 1562.27 - 960

= 602.27 kg

6 0
2 years ago
At its peak, a tornado is 53.0 m in diameter and carries 465-km/h winds. What is its angular velocity in revolutions per second?
expeople1 [14]

Answer:

0.7757 rev/s

Explanation:

d = Diameter of the tornado = 53 m

r = Radius of the tornado = 53/2 = 26.5 m

v = Velocity of wind = 465 km/h

Converting velocity to m/s

465=465\times \frac{1000}{3600}=\frac{775}{6}

Angular velocity

\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{\frac{775}{6}}{26.5}\\\Rightarrow \omega=4.87\ rad/s

\omega=4.87\frac{1}{2\pi}=0.7757\ rev/s

∴ Angular velocity is 0.7757 rev/s

8 0
3 years ago
Question 1
valentina_108 [34]

The distance covered is simply the length of the entire trip, which is 12m + 16m, or 28m.

The displacement is the distance from the starting point to the ending point along with the direction of the net motion. The dog walks 12m east then 16m west, so its resultant displacement is 4m west.

6 0
2 years ago
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