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Temka [501]
3 years ago
7

Explain how gamma rays are produced by energy

Physics
1 answer:
sashaice [31]3 years ago
8 0
Gamma rays<span> are </span>produced<span> in the disintegration of radioactive atomic nuclei and in the decay of certain subatomic particles. When an unstable atomic nucleus decays into a more stable nucleus, the “daughter” nucleus is sometimes </span>produced<span> in an excited state.</span>
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A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r
kykrilka [37]

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

the kinetic energy is given as:

K.E = K.E=\frac{1}{2}I\omega^2

where,

I is the moment of inertia = mr²

on substituting the values, we get

K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2

or

K.E = 0.0075 J

3 0
3 years ago
7. A 1000 kg car is rolling down the street at 2.5 m/s. How fast would a 2500 kg car have to
babunello [35]

1 m/s

Explanation:

To solve this question we use the following formula:

momentum = mass × velocity

momentum of the first car = 1000 kg × 2.5 m/s

momentum of the second car = 2500 kg × X m/s

To bring the cars at rest the momentum of the first car have to be equal to the momentul of the second car.

momentum of the first car = momentum of the second car

1000 kg × 25 m/s = 2500 kg × X m/s

X (velocity of the second car) = (1000 × 25) / 2500 = 1 m/s

Learn more about:

momentum

brainly.com/question/13378780

#learnwithBrainly

7 0
3 years ago
What are two different examples of positive exeleration
sdas [7]

Answer:Accelerating your car to get up to speed on a freeway

An airplane accelerating to take off

Accelerating out of the starting blocks at a track meet

Explanation:

5 0
3 years ago
The cable of a crane is lifting a 950 kg girder. The girder increases its speed from 0.25 m/s to 0.75 m/sin a distance of 5.5m
Umnica [9.8K]

Explanation:

a) How much work is done by gravity?

  • w = f x d
  • w = 950 x 10 x 5.5 = 52250j

b) How much work is done by tension?

  • v²=u²+2as
  • 0.75²=0.25²+2a x5.5
  • 0.56=0.06+2a x5.5
  • 2a x5.5 = 0.56 - 0.06
  • 2a x 5.5 =0.5
  • 11a=0.5
  • a = 0.5/11 = 0.05m/s²

w = f x d

w = 950 x 0.05 x 5.5 = 261.25j

7 0
2 years ago
In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula, we find:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s

5 0
3 years ago
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