Answer:
because they are same and their properties
C
Atomic radius is the distance between the center of the nucleus to the outermost orbital shell of the atom. Assume the atom is like a football stadium and the nucleus of the atom is a ball placed at the center of the pitch. The atomic radius is from the center of the ball to the edge of the football stadium.
Explanation:
This atomic radius decreases from left to right of a periodic table because of increases in protons in the nucleus along the periodic table. This increased proton count has a higher attractive force on the electron orbitals of the atom. This decreases the atomic radius
The atomic radius of atoms down a column of the periodic table increase because an extra orbital shell is added to the atoms with every period down the column.
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The object has been moved by 5 m
Explanation:
The work done by a force when moving an object (which is equal to the energy used to move the object) is given by
where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
In this problem, we have
W = 35 J is the work done
F = 7.0 N is the magnitude of the force
, assuming the force is applied parallel to the direction of motion of the object
Therefore, we can solve the formula for d to find the displacement of the object:
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Answer:
The frequency heard by the motorist is 4313.2 Hz.
Explanation:
let f1 be the frequency emited by the police car and f2 be the frequency heard by the motorist, let v1 be the speed of the police car and v2 be the speed of the motorist and v = 343 m/s be the speed of sound.
because the police car is moving towards the motorist at a higher speed, then the motorist will hear a increasing frequency and according to Dopper effect, that frequency is given by:
f1 = [(v + v2/(v - v1))]×(f2)
= [( 343 + 30)/(343 - 36)]×(3550)
= 4313.2 Hz
Therefore, the frequency heard by the motorist is 4313.2 Hz.
Answer:
670400 J
Explanation:
m = mass of water = 2 kg
T₀ = initial temperature of water = 20 ºC
T = initial temperature of water = 100 ºC
c = heat capacity of water = 4190 J /kg ºC
Q = energy needed to raise the temperature of water
Energy needed to raise the temperature of water is given as
Q = m c (T - T₀)
Inserting the values
Q = (2) (4190) (100 - 20)
Q = 670400 J
