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3 years ago

What could happen if a black whole eat earth

Physics

1 answer:

hram777 [196]3 years ago

7 0

We would all die and theres no way of coming out of a black hole. What's in it is unknown.

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An ice skater has a moment of inertia of 5.0 kg · m2when her arms are outstretched, and at this time she is spinning at 3.0 rev/

KatRina [158]

Answer:

Explanation:

Given

Initial Moment of Inertia $I_1=5 kg-m^2$

initial Spin $N_1=3 rev/s$

$\omega _1=2\pi N_1=2\pi 3=6\pi rad/s$

Final Moment Moment of Inertia $I_2=2 kg-m^2$

Conserving Angular momentum

$L_1=L_2$

$I_1\omega _1=I_2\omega _2$

$5\times 6\pi=2\times \omega _2$

$\omega _2=15\pi rad/s$

$N_2=\frac{\omega _2}{2\pi }$

$N_2=\frac{15\pi }{2\pi}=7.5 rev/s$

8 0

3 years ago

2. Which is longer? အ O 1 mile O 2 kilometers O 1000 mL O 850g

anastassius [24]

Answer:

•2 kilometres.................

5 0

3 years ago

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$

Where in this case:

$m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m$

Therefore:

$E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J$

8 0

3 years ago

If I wanted to generate a maximum emf of 20 V, what angular velocity (radians/sec, aka Hz) would be required given a circular co

sergij07 [2.7K]

Answer:

Angular velocity, $\omega=35.36\ rad/s$

Explanation:

It is given that,

Maximum emf generated in the coil, $\epsilon=20\ V$

Diameter of the coil, d = 40 cm

Radius of the coil, r = 20 cm = 0.2 m

Number of turns in the coil, N = 500

Magnetic field in the coil, $B=9\times 10^{-3}\ T$

The angle between the area vector and the magnet field vector varies from 0 to 2 π radians. The formula for the maximum emf generated in the coil is given by :

$\epsilon=NBA\omega$