Answer:
50N increasing
Explanation:
it looks like we want the net force. this is the sum of all forces acting on the object. here the problem is extremely simplified and we are even given the values of force. this looks like the rocket is producing an upward force of 160 newtons. it also looks like maybe weight (gravity) and drag (air resistance) are producing downward force against the rocket. we can say those would be subtracting from the amount of upwards force. so 160-40-70=50N. since the force is positive that means we will still have a magnitude of force vector that points upward and so we will increase.
Answer:
Explanation:
Initially no of atoms of A = N₀(A)
Initially no of atoms of B = N₀(B)
5 X N₀(A) = N₀(B)
N = N₀ 
N is no of atoms after time t , λ is decay constant and t is time .
For A
N(A) = N(A)₀ 
For B
N(B) = N(B)₀ 
N(A) = N(B) , for t = 2 h
N(A)₀ = N(B)₀
N(A)₀ = 5 x N₀(A)
= 5
= 5 
half life = .693 / λ
For A
.77 = .693 / λ₁
λ₁ = .9 h⁻¹
= 5
Putting t = 2 h , λ₁ = .9 h⁻¹
= 5 
= 30.25
2 x λ₂ = 3.41
λ₂ = 1.7047
Half life of B = .693 / 1.7047
= .4065 hours .
= .41 hours .
Weight can be explained as the force with which the gravity pulls an object. Your weight will not be the same in all planets. In moon, you will weigh far lesser than how much you weigh on the earth. However, in earth and in the moon, your mass will remain the same.
Answer:
a)
b)
.
Explanation:
Given that
Boyle's law
P V = Constant ,at constant temperature
a)
Given that
We know that for PV=C
Now by putting the values
PV= 50 x 0.106
Where P is in KPa and V is in 
b)
PV= C
Take ln both sides
So 
lnP + lnV =lnC ( C is constant)
By differentiating
So
When P= 50 KPa
It indicates the slope of PV=C curve.
It unit is 
.
Or we can say that .
Answer:
Gravity
Explanation:
When the ball is falling to the ground, it is already detached from the table, so the table does not exert any force on it.
Gravity is always present, therefore it is acting on the ball (acting downward), so it must be included into the free-body diagram. Apart from that, there are no other forces acting on the ball (if we neglect air resistance, which is negligible, and it is not mentioned in the options given), therefore the only force which has to be included in the diagram is gravity.
