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Natalka [10]
3 years ago
7

E=mc2 sun is losing 4.3 billion kg mass each second ? can you let me know why ?

Physics
1 answer:
Vanyuwa [196]3 years ago
7 0
Yes I can. You've come to the right place. The reason is because that's where the sun's energy comes from. The sun converts that much mass into energy every second. Then the energy leaves the Sun and shines out into space. A tiny tiny tiny tiny tiny bit of it hits the Earth and it sustains life on Earth.
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What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af
sladkih [1.3K]

Answer:

L=0.0045\ kg-m^2/s

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, \omega=4300\ rpm=450.29\ rad/s

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

L=I\omega

Where I is moment of inertia

For sphere, I=\dfrac{2}{5}mr^2

L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s

So, the magnitude of the angular momentum of the sphere is 0.0045\ kg-m^2/s.

4 0
2 years ago
How much work is required to move
Nikitich [7]
The work is path independent since we have a conservative force.

Thus W=d\cdot\frac{q\cdot U}{d^2}=\frac{3.0\cdot9.0}{0.010}=\boxed{2700 J}

Answer (1)
6 0
3 years ago
A 1kw electric heater is switched on for ten minutes. how much heat does it produce​
Ainat [17]

H= P × t

1kW= 1000w

10 min = 600s

H= 1000×600=600,000J

=> 143.40kcal

4 0
3 years ago
To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit
DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

R = 350ft

a_t = 1.1ft/s^2

PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

27.5625 = 1.21 + \frac{v^4}{122500}

v=42.3877ft/s

Now calculate the angular velocity of the motorcycle

v = r\omega

42.3877 = 350\omega

\omega = 0.1211rad/s

Calculate the angular acceleration of the motorcycle

a_t = r\alpha

1.1 = 350\alpha

\alpha = 3.1428*10^{-3}rad/s^2

Calculate the time needed by the motorcycle to reach an acceleration of

5.25ft/s^2

\omega = \alpha t

0.1211 = 3.1428*10^{-3}t

t = 38.53s

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

a = 1.7187ft/s^2

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

3 0
3 years ago
How much does a 10 kg
Goshia [24]

Answer:

0

Explanation:

because during free fall we experience wieghtlessness

5 0
1 year ago
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