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3 years ago

In nature, what gets energy from nuclear fusion?

Physics

1 answer:

Andru [333]3 years ago

7 0

Answer:

Fusion powers the Sun and stars as hydrogen atoms fuse together to form helium, and matter is converted into energy.

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How is sound produced

Dafna11 [192]

Sounds are created when something vibrates (shakes back and forth), sending waves of vibrations into the ears of the listener. When a bell is struck, the metal vibrates. The vibrations travel through the air as sound waves.

hope this helps

7 0

2 years ago

Mechanical advantage of a machine can be increased by designing it for:

Tems11 [23]

Answer:(4).

Explanation:

8 0

3 years ago

A weather balloon is designed to expand to a maximum radius of 24 m at its working altitude, where the air pressure is 0.030 atm

nlexa [21]

Answer:

Radius at liftoff 8.98 m

Explanation:

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

First, we assume balloon is spherical in nature,

and that the working gas obeys the gas laws.

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere = $\frac{4}{3} \pi r^{3}$

volume of balloon = $\frac{4}{3}$ x 3.142 x $24^{3}$ = 57913.34 m^3

using the gas equation,

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.

imputing values, we have

$\frac{0.03*57913.34}{200}$ = $\frac{1*V2}{349}$

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = 3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.

from the formula volume of a sphere,

V = $\frac{4}{3} \pi r^{3}$ = $\frac{4}{3}$ x 3.142 x $r^{3}$ = 3031.76

4.19 $r^{3}$ = 3031.76

$r^{3}$ = 3031.76/4.19

radius r of the balloon on liftoff = $\sqrt[3]{723.57}$ = 8.98 m

4 0

3 years ago

An inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extra

WARRIOR [948]

Answer:

The inventors claim is not real

a) No the the freezer cannot operate in such conditions

Explanation:

From the question we are told that

The power input is $P_i = 0.25 kW = 0.25 *10^{3} \ W$

The rate of heat transfer $J = 3050 J/s$

The temperature of the freezer content is $T = 270 \ K$

The ambient temperature is $T_a = 293 \ K$

Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as

$COP = \frac{T }{Ta - T}$

substituting values

$COP = \frac{270 }{293 - 270}$

$COP =11.7$

Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as

$COP = \frac{J}{P_i}$

substituting values

$COP = \frac{3050}{0.25 *10^{3}}$

$COP = 12.2$

Now given that the COP of an ideal refrigerator is less that that of a real refrigerator then the claims of the inventor is rejected

This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP

4 0

3 years ago

A satellite orbiting Earth has a tangential velocity of 5000 m/s. Earth’s mass is 6 × 1024 kg and its radius is 6.4 × 106 m.

Aloiza [94]

When a satellite is moving around the Earth's orbit, two equal forces are acting on it. The centripetal and the centrifugal force. The centripetal force is the force that attracts the object toward the center of the axis of rotation. The opposite force is the centrifugal force. It draws the object away from the center. When these forces are equal, the satellite uniformly rotates along the orbit.

Centripetal force = Centrifugal force
Mass of satellite * centripetal acceleration = Mass of satellite * centrifugal acceleration
Centripetal acceleration = Centrifugal acceleration

$\frac{ M_{E}*G }{ ( r_{E} )^{2} } =$ ω^2r

where

$M_{E}$ = mass of earth
G = gravitational constant = 6.6742 x 10-11 m3 s-2 kg-1
 $r_{E}$ = radius of earth
ω = angular velocity
r = radius of orbit

To convert to angular velocity:

Tangential velocity = rω
ω = 5000/r

Then,

$\frac{ (6 \ x \ 10^{24}) *(6.6742 \ x \ 10^{-11} ) }{ ( 6.4 \ x \ 10^{6})^{2} }= ( \frac{5000}{r} )^{2} r$

r = 2557110.465 m

Therefore, the distance of the centers of the earth and the satellite is 2.6 x 10^6 m.

4 0

3 years ago

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