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3 years ago

Find the kinetic energy of a bullet weighing 0.10 N if its velocity is 2.0 X 103 m/s?

Physics

1 answer:

irina1246 [14]3 years ago

5 0

The kinetic energy of the bullet is 20.4 kJ.

<u>Explanation:</u>

Kinetic energy of a bullet will be equal to the product of mass of the bullet with the square of velocity or speed of the bullet and then the half of that product value.

But here the mass of the bullet is not given, instead the weight of the bullet is given in terms of force. So from this, we have to first find the mass of the bullet.

We know that as per Newton's second law of motion, force is directly proportional to the product of mass and acceleration. So here the acceleration will be equal to the acceleration due to gravity as it is weight of the object.

So F = mg

0.10 N = m × 9.8

So ,the mass of the bullet is 0.0102 kg.

Now, we know the mass and velocity of the bullet is given as 2000 m/s.

So,

kinetic energy = $\frac{1}{2}$ × m × v²

kinetic energy = 0.5 × 0.0102 × 2000 × 2000 = 20.4 kJ

Thus, the kinetic energy of the bullet is 20.4 kJ.

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C- Sigma Waves, This is what my teacher used to get us to remember https://www.youtube.com/watch?v=bjOGNVH3D4Y

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3 years ago

A sodium atom will absorb light with a wavelength near 589 nm if the light is within 10 MHz of the resonant frequency. The atomi

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Answer:

i)20369 photons

ii) 40 ps

Explanation:

Momentum of one Sodium atom:

$P=m*v =600m/s*23amu*\frac{1 kg}{6.02*10^{23}amu}\\P=2.29*10^{-23}kgm/s$

In other to stop it, it must absorb the same momentum in photons:

$P=2.29*10^{-23}kgm/s=n_{photons}*\frac{h_{planck}}{\lambda}\\=n*\frac{6.63*10^{-34}}{589*10^{-9}} \\==>n=20369 photons$

Now, for the minimun time, we use the speed of light and the wavelength. For the n photons:

$t=n*T=n*\frac{\lambda}{c} =20369*\frac{589nm}{3*10^{8}m/s}=4*10^{-11} second=40 ps$

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3 years ago

The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha

katovenus [111]

Answer:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.

While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.

And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.

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3 years ago

Read 2 more answers

Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the

kifflom [539]

Answer:

Height will be 3.8971 m

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity $\omega =4.46rad/sec$

Moment of inertia is given by $I=\frac{1}{2}mr^2$

Transnational Kinetic energy is given by $KE=\frac{1}{2}mv^2$ as we know that v = $v=\omega r$

So $KE=\frac{1}{2}m(\omega r)^2$

Rotational kinetic energy is given by $KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2$

Potential energy is given by mgh

According to energy conservation

$mgh=\frac{1}{2}m(\omega r)^2+\frac{1}{4}m(\omega r)^2$

$h=\frac{3r^2\omega ^2}{4g}=\frac{3\times 1.60^2\times 4.46^2}{4\times 9.8}=3.8971m$

3 0

3 years ago

The wheels of an automobile are locked as it slides to a stop from an initial speed of 30.0 m/s. If the coefficient of kinetic f

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Answer:

x = 76.5 m

Explanation:

Let's use Newton's second law at the point of contact between the wheel and the floor.

fr = m a

fr = miy N

N-W = 0

N = W

μ mg = m a

a = miu g

a = 0.600 9.8

a = 5.88 m / s²

Having the acceleration we can use the kinematic relationships to find the distance

$v_{f}$ ² = v₀² + 2 a x

$v_{f}$ = 0

x = -v₀² / 2 a

Acceleration opposes the movement by which negative

x = - 30²/2 (-5.88)

x = 76.5 m

8 0

3 years ago