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Ganezh [65]
3 years ago
12

You serve a volleyball with a mass of 0.77 kg. The ball leaves your hand with a velocity of

Physics
1 answer:
dimaraw [331]3 years ago
7 0
<h2>Given :</h2>

  • mass (m) = 0.77 kg

  • velocity (v) = 2.33 m/s

<h2>Solution :</h2>

\boxed{ \mathrm{Kinetic  \: Energy =  \frac{1}{2} m {v}^{2} }}

  • \dfrac{1}{2}  \times 0.77 \times (2.33) {}^{2}

  • \dfrac{418.0253}{2}

  • 209.01265  \: \: joules

  • \mathrm{209.01  \: \: joules} \:  \:  \: (approx)

_____________________________

\mathrm{ \#TeeNForeveR}

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I NEED HELP PLEASE, THANKS! :)
makkiz [27]

Answer:

They move outwards.

They don't intersect each other at any point.

They show the electric field.

Explanation:

8 0
3 years ago
Read 2 more answers
You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m
slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock m=500 gm

diameter of circle d=2.2 m

radius r=\frac{2.2}{2}=1.1 m

At highest Point

mg+N=\frac{mv^2}{r}

At highest Point N=0 because mass is just balanced by centripetal Force

thus mg=\frac{mv^2}{r}

v=\sqrt{gr}

v=\sqrt{9.8\times 1.1}

v=\sqrt{10.78}

v=3.28 m/s

6 0
3 years ago
An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a
Lapatulllka [165]

Answer:

The speed of the electron is 2.55\times 10^3\ m/s.

Explanation:

Given that,

The magnitude of electric field, E=1.32\ kV/m=1.32\times 10^3\ V/m

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then,  the magnitude of electric field and magnetic field is balanced such that :

evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s

or

v=2.55\times 10^3\ m/s

So, the speed of the electron is 2.55\times 10^3\ m/s. Hence, this is the required solution.

3 0
3 years ago
When a star goes from the main sequence to the red giant phase, it becomes larger rather than smaller. This is true because grav
GuDViN [60]
I believe this is electron degeneracy, because the star is essentially having too many reactions too fast and collapses in on itself eventually.
4 0
3 years ago
The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a
likoan [24]

Answer:

14869817.395 m

Explanation:

\theta=22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians

From Rayleigh Criterion

\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0
3 years ago
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