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3 years ago

PLEASE HELP! ASAP

Physics

2 answers:

hoa [83]3 years ago

8 0

Explanation:

Specifically, small amounts of mass are turned into energy from the breaking up (fission) or combination (fusion) of the nuclei of atoms. Even spontaneous radioactive decay converts a bit of mass into incredible amounts of energy.

Damm [24]3 years ago

5 0

Answer:

A. Uranium breaks apart into krypton, barium, and three neutrons.

Explanation:

did the test !!

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What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the the

timurjin [86]

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = $I_0$

So, M.I. of disk B = $3I_0$

Angular velocity of A = $\omega_0$

So the kinetic energy of the disk A = $\frac{1}{2}I_0\omega^2$

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

$I_0\omega_0 = I_0 \omega + 3I_0 \omega$

$I_0\omega_0 = 4I_0 \omega$

$\omega = \frac{\omega_0}{4}$

Now,

$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$

$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$

$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$

$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$

$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4}$

$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$

$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J$

Therefore, the maximum initial K.E. = 3066.67 J

3 0

3 years ago

Three identical train cars, coupled together, are rolling east at speed v0. A fourth car traveling east at 2v0 catches up with t

aliina [53]

Answer: $v_o$

Explanation:

It is given that three cars has same mass m with speed $v_o$

suppose rest two cars also has same mass m

As there is no external force therefore momentum is conserved

Initial Momentum $P_i$

$P_i=3mv_0+m(2v_0)+m\times 0$

Final momentum $P_f$

$P_f=5m\times v$

where v=final velocity

$P_i=P_f$

$5mv_o=5mv$

$v=v_o$

thus final velocity is $v_o$

8 0

4 years ago

In which position are the earth, moon, and sun during a full moon?

Arte-miy333 [17]

Answer:

The answer is B, although technically that is an eclipse.

5 0

4 years ago

when a seagull picks an oyster up into the sky and then lets it drop on the rocks below to open the shell; where is the oysters

Sati [7]

Answer:

When a seagull picks an oyster up into the sky and then lets it drop on the rocks below to open the shell; where is the oyster's potential energy greatest? Where is its kinetic energy greatest?

Potential energy is greatest at maximum height; kinetic energy is greatest just before the oyster strikes the ground.

Explanation:

5 0

4 years ago

Read 2 more answers

A radio wave has a frequency of 3.20 × 108 hz. what is the energy (in j) of one photon of this radiation? enter your answer in s

sergij07 [2.7K]

The formula to be used in this problem is:E = hν
Where:
ν = frequency in Hz, or s^-1.
From time to time you see E = hf.
h = 6.636 x 10^-34 J.s and is named as Planck's constant.

E = (6.636 x 10^-34 J.s)(3.20 x 10^8 s^-1)= 2.12352 x 10^-25 JThis is the energy of 1 photon.

7 0

4 years ago