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yulyashka [42]
3 years ago
10

During the fission reaction shown, how did the target nucleus change ?

Physics
1 answer:
Zarrin [17]3 years ago
5 0

Answer:

A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

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Vsevolod [243]

Explanation:

- Newton's first law of motion:

"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force

In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.

- Newton's second law of motion:

"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:

F=ma

where F is the net force, m is the mass of the object, and a its acceleration"

In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.

"Newton's third law of motion:

"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.

3 0
3 years ago
A 5.0-kg centrifuge takes 95 s to spin up from rest to its final angular speed with constant angular acceleration. A point locat
stellarik [79]

Answer:

(a) 17.37 rad/s^2

(b) 12479

Explanation:

t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0

w = v / r = 99 / 0.06 = 1650 rad/s

(a) Use first equation of motion for rotational motion

w = w0 + α t

1650 = 0 + α x 95

α = 17.37 rad/s^2

(b) Let θ be the angular displacement

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 α θ

1650^2 = 0 + 2 x 17.37 x θ

θ = 78367.87 rad

number of revolutions, n = θ / 2 π

n = 78367.87 / ( 2 x 3.14)

n = 12478.9 ≈ 12479

4 0
3 years ago
physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin
vova2212 [387]

Answer: Vb is the vector  (-5.37m/s,  8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s,  8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

7 0
3 years ago
A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.
Gala2k [10]
It’s not in English can you change it please
3 0
2 years ago
a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature
igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber V=1 m^3

Initial Temperature T_1=300 K

Final Temperature T_2=400 K

Heat Supplied Q=10 J

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

\Delta U=Q-W

as the volume is fixed therefore work

W=\int PdV=0

thus \Delta U=mc_v\Delta T=Q

c_v for mono-atomic gas is 12.471 J/K-mol

n\times 12.471\times (400-300)=10

n=0.008018 mol

and 1 mole contains 6.022\times 10^{23} molecules

thus  No of molecules=0.008018\times 6.022\times 10^{23}

No of molecules=4.82\times 10^{21} molecules

3 0
3 years ago
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