Initial volume of mercury is
V = 0.1 cm³
The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.
Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³
The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
= (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
= 4.5 cm
Answer: 4.5 cm
F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
So since there is half times the gravity on this unknown planet that has twice earth's mass and twice it's radius, then the person can jump twice as high. 1.5*2= 3m high
Answer:
2. at the lowest point
Explanation:
The motion of the pendulum is a continuous conversion between kinetic energy (KE) and gravitational potential energy (GPE). This is because the mechanical energy of the pendulum, which is sum of KE and GPE, is constant:
E = KE + GPE = const.
Therefore, when KE is maximum, GPE is minimum, and viceversa.
So, the point of the motion where the KE is maximum is where the GPE is minimum: and since the GPE is directly proportional to the heigth of the bob:
we see that GPE is minimum when the bob is at the lowest point,so the correct answer is
2. at the lowest point
Answer:
C
Explanation:
Vector A points up
Vector B points right
The combination must be both up and right which is C
Answer:
If child weight is equal to rope force then child will move with uniform speed
or we can say that the child will remain at rest in his position
Explanation:
As we know that child is hanging by rope
so here there will be two forces on the child
1) Weight or gravitational force which act vertically downwards
2) Tension in the rope which act vertically upwards
Now if child will accelerate upwards then tension force must be more than the weight of the child
If tension force is less than the weight then child will decelerate and his speed will decrease
if tension force is equal to child weight then in that case the child will remain at rest or it will move with same speed
