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Rashid [163]
3 years ago
10

Why does a rolling sphere slows down?​

Physics
1 answer:
Iteru [2.4K]3 years ago
4 0
There’s frictional force acting on the sphere, which causes it to gradually slow down, and eventually come to a stop.
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1 gram of radium is reduced by 3.1 mg in 5 years by alpha decay.
Whitepunk [10]

Answer:

1 gm = 1000 mg

N = No e^-y t       where y = lambda the decay constant

N / N0 = .9969

ln .9969 = - y t

-.00310 = - 5 y

y = .00310 / 5 = .000621

So the half-life = .693 / y

T^1/2 = .693 / .000621 = 1116 or about 1120 yrs

4 0
3 years ago
Explain in detail the birth life and death of a main sequence star give an example
galina1969 [7]
Birth Star:
Main sequence stars fuse hydrogen atoms to form helium atoms in their cores. About 90 percent of the stars in the universe, including the sun, are main sequence stars. These stars can range from about a tenth of the mass of the sun to up to 200 times as massive. Stars start their lives as clouds of dust and gas.
Death Star:
Dense regions in the clouds collapse and form "protostars". Initially, the gravitational energy of the collapsing star is the source of its energy. Once the star contracts enough that its central core can burn hydrogen to helium, it becomes a "main sequence" star.
8 0
3 years ago
How many atoms are in a 4.30 cm à 4.30 cm à 4.30 cm cube of aluminum?
Elan Coil [88]
Hmm that's a tuff one let me think
4 0
3 years ago
What is a analogy for transition metals
Ganezh [65]

<em>A simple metallic band model is proposed for the transition metal mono antimonides, by analogy to the transition metals.</em>

6 0
3 years ago
An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis
Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (\Delta K) and part is lost because of the air resistance (E_{lost}).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

where

v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

8 0
3 years ago
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