Answer:
Explanation:
Given
mass of car 
Initial velocity of car 
towards east
Time taken to stop 
Force exerted 
change in momentum is given by impulse imparted to the car
negative Sign indicates that impulse is imparted opposite to the direction of motion
magnitude of momentum 
Answer:
Elements in the same period have the same number of electron shells; moving across a period (so progressing from group to group), elements gain electrons and protons and become less metallic. This arrangement reflects the periodic recurrence of similar properties as the atomic number increases.
Explanation:
The Periodic Table can predict the properties of new elements, because it organizes the elements according to their atomic numbers. ... They hope that the two nuclei at the centre of these atoms will fuse and form a heavier nucleus. When these heavy elements form, they are usually highly unstable.
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
Energy can be one answer! There are many, but energy is a main one.
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
