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creativ13 [48]
3 years ago
14

The Mass of the Sun is 1.989 × 10^30 kg. The speed of light is 3 x 10^8 m/s. Assuming the Sun is 75% hydrogen, how much Energy c

ould be produced if all of the hydrogen is converted into energy? Use Einstein's famous equation E = mc^2 where E will be in Joules.
Physics
1 answer:
bogdanovich [222]3 years ago
8 0

Answer:

E = 13.428 \times 10^{46} J

Explanation:

Mass of the Sun = 1.989 \times 10^{30} Kg

Amount of Hydrogen = 75% of Mass of Sun,

thus mass of Hydrogen (m) in the Sun is,

m = 1.989 \times 10^{30} \times \frac{75}{100}

m = 1.492 \times 10^{30}

Speed of Light (c) = 3 \times 10^{8} m/s

Thus, energy(E) produced if whole of hydrogen is converted into energy,

E = mc²

E = 1.492 \times 10^{30} \times (3 \times 10^{8} )^{2}

E = 1.492 \times 9 \times 10^{46}

E = 13.428 \times 10^{46} J

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Answer:

wavelenght

Explanation:

The wavelength is the spatial period of a wave, analogous to the temporal period, it is the distance between two consecutive points with maximum amplitude that are repeated in space . In the waves of the sea, the wavelength is easily observed in the separation between two consecutive ridges.

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3 years ago
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Students in a lab are doing experiments involving the motion of objects. They pull an object across a table and use
Svetlanka [38]

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Explanation:

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3 years ago
Object 1 of mass m moves with speed v in the positive direction. Object 2 of mass 3 m moves with speed 4 v in the negative x-dir
Klio2033 [76]

Answer:

This means that the kinetic energy of second object is 48times that of the first object

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion e.g motion of an accelerating car. Mathematically,

Kinetic energy = 1/2mv² where;

m is the mass of the object

v is the velocity of the object

If Object 1 of mass m moves with speed v in the positive direction, its kinetic energy will be expressed as;

K1 = 1/2mv²

For Object 2 of mass 3m moving with speed 4v in the negative x-direction, its kinetic energy can be expressed as;

K2 = 1/2(3m)(4v)²

K2 = 1/2(3m)(16v²)

K2 = (3m)(8v²)

K2 = 24mv²

To compare the kinetic energy of both bodies, we will take the ratio of K2:K1 to have;

K2/K1 = 24mv²/(1/2)mv²

K2/K1 = 24/(1/2)

K2/K1 = 48

K2 = 48K1

This means that the kinetic energy of second object is 48times that of the first object and moving in the negative x direction since the body of mass 3m initially moves in the negative x direction.

3 0
4 years ago
.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric
maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0
3 years ago
A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an
agasfer [191]

Answer:

t=\sqrt{h/g}

Explanation:

We use the kinematics equation to solve this question:

y(t)=y_{o}+v_{o}t+1/2*a*t^{2}

v_{o}=0    because the ball is dropped

a=-g         the acceleration is the gravity, negative because it points downwards

y_{o}=h     initial height

y(t)=h/2     final height

So:

h/2=h-1/2*g*t^{2}

t=\sqrt{h/g}

8 0
3 years ago
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