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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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Riley is cleaning his room(a once a year activity) and pulls the Hoover with a force of 8 N a total distance of 20 metres. How m

Radda [10]

Answer:

160J

Explanation:

Given force = 8N and total distance = 20 meters

Workdone = force x distance

= 8 x 20

= 160J

Therefore, workdone by Riley in pulling the hoover is 160J

7 0

3 years ago

The Balmer series in hydrogen atom produces only infrared emission. O True O False

lorasvet [3.4K]

Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

$n_{i}=2$ .

$[tex]n_{f}=3,4,5,.....\infty$ .

Now the value of wavelength can be calculated as,

$\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}$ .

Here, $R=109677 cm^{-1}$ .

And $n_{f}=3$ .

Now,

$\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3})$ .

Therefore,

$\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm$

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

8 0

3 years ago

Read 2 more answers

The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the locat

Nonamiya [84]

Answer:

E=0

Explanation:

The electric field at the centre of the shell is zero because total enclosed charge in the nucleus is zero

7 0

3 years ago

T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?

Debora [2.8K]

Answer:

$P_J=195N$

Explanation:

From the question we are told that

$Mass\ of T-joe\ M_J=65\\Velocity\ of T-joe\ V_J=3m/s\\Mass of\ T-Brud\ M_B=50kg\\Velocity\ of T-Brud\ V_B=3m/s\\$

Generally the equation for momentum is mathematically given by

$P=mv$

Therefore

T-Joe momentum $P_J$

$P_J=65*3$

$P_J=195N$

5 0

2 years ago

Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest

mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

v = 9.603 m/s

6 0

3 years ago