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Dafna1 [17]
4 years ago
7

How much mass does the sun lose through nuclear fusion per second?

Physics
1 answer:
Mekhanik [1.2K]4 years ago
3 0

Answer:

1.10^6 kg of mass per second

Explanation:

All energy lost by the sun comes from nuclear fusion.

Sun loses energy at 2.5*10^{19}J per hour, that is 9*10^{22}J/s

To find the mass lost by the sun in liberation of energy you use the famous Einstein's equation:

E=mc^2\\\\m=\frac{E}{c^2}=\frac{9*10^{22}}{(3*10^{8}m/s)^2}=1*10^6kg

hence, the sun liberates 1.10^6 kg of mass per second

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An Arrow (0.5 kg) travels with velocity 60 m/s to the right when it pierces an apple (1 kg) which is initially at rest. After th
Rom4ik [11]

Answer:

Velocity after collision will be 20 m/sec

Explanation:

We have given mass of arrow m_1=0.5kg

Mass of arrow v_1=60m/sec

Mass of apple m_2=1kg

Apple is at rest so velocity of apple v_2=0m/sec

According to conservation of momentum

Momentum before collision is equal to momentum after collision

m_1v_1+m_2v_2=(m_1+m_2)v

0.5\times 60+1\times 0=(0.5+1)v

1.5v=30

v = 20 m/sec

5 0
4 years ago
Easy quiz easy points 7
Darya [45]
The answer is 2: hush
5 0
3 years ago
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You push a 10 n object 10 meters. How much work was done on the object
Contact [7]
Work= Force * Distance
Work= 10N * 10m
Work= 100J
3 0
3 years ago
If a car accelerates at a uniform 4.0 m/s2, how long will it take to reach a speed of 80 km/hr, starting from rest?.
Mila [183]

Answer:

5.6s

Explanation:

<u>Step1:</u> Identify the given parameters

acceleration(a) = 4m/s²

initial velocity (u)= 0 (since the car started from rest)

final velocity (v) = 80km/hr

final velocity(v) in m/s = 80\frac{(km)}{(hr)} X\frac{(1000m)}{(1km)} X\frac{(1 hr)}{(3600s)}

= \frac{80,000(m)}{3,600(s)}

=22.22m/s

<u>Step2:</u> calculate the time it takes the car to attain speed of 80km/hr = 22.22m/s

acceleration(a) = change in velocity per time

acceleration = \frac{(final velocity)-(initial velocity)}{time}

a= \frac{v-u}{t}

t=\frac{v-u}{a}

t=\frac{22.22-0}{4} (s)

a= 5.555 seconds

a≅5.6s

4 0
4 years ago
En la figura, la tensión desarrollada en cada
malfutka [58]

Answer:

Parte A

El ángulo con respecto al horizonte, de la segunda cuerda es de aproximadamente 19,47°

Parte B

La masa de la caja que se va a sostener es de aproximadamente 0,7808 kg.

Explanation:

Parte A

Los parámetros dados son;

La tensión en la cuerda, T₁ = 8 N

La tensión en la cuerda, T₂ = 6 N

El ángulo de inclinación de la primera cuerda con la horizontal, θ₁ = 45°

Sea θ₂ el ángulo de inclinación de la segunda cuerda, obtenemos;

T₁·cos (θ₁) = T₂·cos (θ₂)

∴ 8 N × cos (45°) = 6 N × cos (θ₂)

cos (θ₂) = 8 N × cos (45°) / (6 N) = (√2)/2 × (4/3) = (2·√2)/3

θ₂ = arcos ((2·√2) / 3) ≈ 19,47°

El ángulo con respecto al horizonte, de la segunda cuerda, θ₂ ≈ 19,47°

Parte B

El peso de la caja, W = T₁·sin (θ₁) + T₂·sin (θ₂)

∴ W = 8 N × sen (45 °) + 6 N × sen (19,47 °) ≈ 7,66 N

El peso de la caja que se va a sostener, W ≈ 7,66 N

La masa de la caja que se va a sostener, m ≈ 7,66 N / (9,81 m/s²) ≈ 0,7808 kg

8 0
3 years ago
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