An object rolls East at a steady speed of 12 m/s for 3 seconds. What distance did the object travel

Two electrons travel towards each other at 0.2 c parallel to the laboratory x-axis. What is the relative velocity of one electro

Leya [2.2K]

1) In the reference frame of one electron: 0.38c

To find the relative velocity of one electron with respect to the other, we must use the following formula:

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

where

u is the velocity of one electron

v is the velocity of the second electron

c is the speed of light

In this problem:

u = 0.2c

v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)

Substituting, we find:

$u'=\frac{0.2c+0.2c}{1+\frac{(0.2c)(0.2c)}{c^2}}=\frac{0.4c}{1+0.04}=0.38c$

2) In the reference frame of the laboratory: -0.2c and +0.2c

In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are

+0.2c

-0.2c

5 0

3 years ago

The first scientist to show that atoms emit any negative particles was

densk [106]

The first scientist to show that atoms emit any negative particles was : J.J Thomson

4 0

3 years ago

What effect does time have on the speed of a moving object

sergejj [24]

Velocity is d/t distance over time. Increase velocity (speed) decrease. Increase d velocity increases.

6 0

3 years ago

"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at

MrRissso [65]

Answer:

life (N) of the specimen is 117000 cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = $\frac{(f\times Su)^2}{Se}$ ......................1

put here value and we get

a = $\frac{(0.82\times 120)^2}{60}$

a = 161.4 kpsi

so coefficient value (b) will be

b = $-\frac{1}{3}log\frac{(f\times Su)}{Se}$

b = $-\frac{1}{3}log\frac{(0.82\times 120)}{60}$

b = −0.0716

so here number of cycle N will be

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value and we get

N = $(\frac{ 70}{161.4})^{1/-0.0716}$

N = 117000

so life (N) of the specimen is 117000 cycles

7 0

3 years ago

At a distance r from a charge e on a particle of mass m the electric field value is

mel-nik [20]

At a distance r from a charge e on a particle of mass m the electric field value is 8.9876 × 10⁹ N·m²/C². Divide the magnitude of the charge by the square of the distance of the charge from the point. Multiply the value from step 1 with Coulomb's constant.

<h3>what is magnitude ?</h3>

Magnitude can be defined as the maximum extent of size and the direction of an object.

It is used as a common factor in vector and scalar quantities, as we know scalar quantities are those quantities that have magnitude only and vector quantities are those quantities have both magnitude and direction.

There are different ways where magnitude is used Magnitude of earthquake, charge on an electron, force, displacement, Magnitude of gravitational force

For more details regarding magnitude, visit

brainly.com/question/28242822

#SPJ1

7 0

1 year ago