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3 years ago

Find the momentum of a particl with a mass of one gram moving with half the speed of light.

Physics

1 answer:

joja [24]3 years ago

6 0

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

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What is the energy range (in joules) of photons of wavelength 410 nm to 750 nm ? Express your answers using two significant figu

andreyandreev [35.5K]

Answer:

4.9 x 10^-19 J, 2.7 x 10^-19 J

Explanation:

first wavelength, λ1 = 410 nm = 410 x 10^-9 m

Second wavelength, λ2 = 750 nm = 750 x 10^-9 m

The relation between the energy and the wavelength is given by

E = h c / λ

Where, h is the Plank's constant and c be the velocity of light.

h = 6.63 x 10^-34 Js

c = 3 x 10^8 m/s

So, energy correspond to first wavelength

E1 = (6.63 x 10^-34 x 3 x 10^8) / (410 x 10^-9) = 4.85 x 10^-19 J

E1 = 4.9 x 10^-19 J

So, energy correspond to second wavelength

E2 = (6.63 x 10^-34 x 3 x 10^8) / (750 x 10^-9) = 2.652 x 10^-19 J

E2 = 2.7 x 10^-19 J

4 0

3 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him

juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal direction

$m_{b}v_{b}\cos\theta= m_{c}v_{c}$

$v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}$

Put the value into the formula

$v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}$

$v_{c}=0.0345\ m/s$

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0

3 years ago

A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su

madreJ [45]

maximum static friction acting on the object will be

$F_s = \mu_s mg$

plug in all values

$F_s = 0.40 \times 15 \times 9.8 = 58.8 N$

So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force

So here it is given that applied force is 20 N

so here object will not move due to this force and it will remain at rest always

due to this applied force

6 0

3 years ago

Immediately after it has started to rain, dust particles and oil residue float on the water and ____________ the traction of tir

Mkey [24]

When it rains, dust particles and oil residues float on the water and this reduces the traction of tires.

<h3>What is traction?</h3>

This concept refers to a force between the tires and road that causes the movement of the wheels or vehicle is slower.

<h3>What happens with traction when it rains?</h3>

It is well-known more accidents occur when it rains, which is caused by cars slidding on the road. This is because when it rains traction or the grip of the wheel drastically reduces.

Learn more about traction in: brainly.com/question/14525337

8 0

2 years ago

3.) An engineer is designing the runway for an airport. Of the planes that will use the airport,

scoray [572]

Answer: 704

Explanation:Vi = 0 m/s

vf = 65 m/s

a = 3 m/s2

d = ??

vf2 = vi2 + 2*a*d

(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d

(4225 m 2/m2)/(6 m/s2) = d

d = 704 m

5 0

3 years ago

Read 2 more answers