Radiation with a wavelength of 238 nm shines on a metal surface and ejects electrons that have a maximum speedof 3.75 X 105 m/s.

A 3.0kg weight W is initially at rest on incline AB, which is raised 40° above the horizontal. The effective coefficient is

lakkis [162]

(a) The acceleration of the system is determined as 1.58 m/s².

(b) The relative weight of P is pounds is determined as 0.14 lb.

<h3>Acceleration of the system</h3>

The acceleration of the system is calculated as follows;

W - T = m₂a --- (1)

T = m₁a ----(2)

μmgsinθ - m₁a = m₂a

(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a

5.67 - 0.6a = 3a

5.67 = 3.6a

a = 5.67/3.6

a = 1.58 m/s²

<h3> Relative Weight of P</h3>

W = ma

W = 0.4 x 1.58

W = 0.632 N = 0.14 lb

Learn more about weight here: brainly.com/question/2337612

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3 0

1 year ago

Calculate the electric potential at point A, the middle of the rectangle, and at point B, the middle of the right-hand side of t

dsp73

Answer:

With the help of formula.

Explanation:

We can calculate the electric potential of any point through the formula of electric potential which is given below.

Electric potential = Coulomb constant x charge/ distance of separation.

Symbolically it can be written as, V = k q/ r where

V = electric potential

k = Coulomb constant

q = charge

r = distance of separation

If we have all these data, we can simply put the data in the formula and we will get the value of electric potential.

5 0

2 years ago

The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within

zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, $m=9.1\times 10^{-31}\ kg$

Initial speed of electron, u = 0

Final speed of electrons, $v=3\times 10^7\ m/s$

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(3\times 10^7)^2}{2\times 0.063}$

$a=7.14\times 10^{15}\ m/s^2$

Now we will find the electric field required in the tube as :

$ma=qE$

$E=\dfrac{ma}{q}$

$E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}$

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0

2 years ago

A boat is heading across a river at a velocity of 25 mph. The river is flowing downstream at 10 mph. What is the actual velocity

AnnyKZ [126]

Principle: Pythagorean theorem

Given:
Let a = boat velocity = 25 miles per hour
b = river velocity = 10 miles per hour

Required: c = actual velocity

Solution:

c^2 = a^2 + b^2
c^2 = (25)^2 + (10)^2
c^2 = 725
(c^2)^0.5 = (725)^0.5
c = 26.93 miles per hour

<em>ANSWER: Actual boat velocity is 26.93 miles per hour</em>

8 0

3 years ago

Why does a pencil look bent when it is placed in a glass of water?

Nonamiya [84]

Answer:

This is due to bending of light ray passing through the medium.

Explanation:

This can be explained using the concept of refraction.

Refraction can be defined as the bending of light ray as it travels from one medium to another (it can be from the less dense to denser medium or denser to less dense medium)

For that of pencil in water, light ray travels from less dense medium(surrounding air) to denser medium(water). This causes the pencil to bend away from the normal rather than towards the normal (angle of refraction being greater than angle of incidence) and hence the reason for the bending.

NB: Normal line is perpendicular to the plane between the medium.

7 0

3 years ago