Answer: 0.86 × 10^14
Explanation:
Given the following :
Radius of proton = 1.2 × 10-15 m
Radius of hydrogen atom = 5.3 × 10-11 m
Density of proton could be calculated thus:
Mass of proton = 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45
Density = mass / volume
Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)
= 0.2306 × 10^18
Density of hydrogen atom:
Mass of hydrogen atom= 1.67 × 10^-27 kg
Using the formula :
(4/3) × pi × r^3
(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31
Density = mass / volume
Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)
= 0.2676 × 10^4
Ratio is thus:
Density of proton / density of hydrogen atom
0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14
Answer:
Option A
Carrots are cut into small pieces and mixed into a salad
Explanation:
When physical changes occur, the actual composition remain the same but the molecules are re-arranged. Therefore, when carrots are cut into smaller pieces and mixed into salad, there will be no chemical reaction hence the actual composition will remain the same despite being cut and molecules in it re-arranged. Considering the other options, new substances are formed hence they are deemed as chemical changes. Therefore, option A is correct.
Answer:
Explanation:
Situations in which an electron will be affected by an external electric field but will not be affected by an external magnetic field
a ) When an electron is stationary in the electric field and magnetic field , he will be affected by electric field but not by magnetic field. Magnetic field can exert force only on mobile charges.
b ) When the electron is moving parallel to electric field and magnetic field . In this case also electric field will exert force on electron but magnetic field field will not exert force on electrons . Magnetic field can exert force only on the perpendicular component of the velocity of charged particles.
Situations when electron is affected by an external magnetic field but not by an external electric field
There is no such situation in which electric field will not affect an electron . It will always affect an electron .
The speed of the ball just before impact was v=√(2gh) = 6.26m/s. The acceleration is twice this over the time (twice because the second speed is the same in the other direction, meaning the total change in speed is 2V)
a = 12.52/0.10 = 125.2m/s²
The force is F=ma, so F = 0.5kg·125.2m/s² = 62.6N
Answer:
The coefficient of static friction between the box and floor is, μ = 0.061
Explanation:
Given data,
The mass of the box, m = 50 kg
The force exerted by the person, F = 50 N
The time period of motion, t = 10 s
The frictional force acting on the box, f = 30 N
The normal force on the box, η = mg
= 50 x 9.8
= 490 N
The coefficient of friction,
μ = f/ η
= 30 / 490
= 0.061
Hence, the coefficient of static friction between the box and floor is, μ = 0.061
