Answer:

Explanation:
We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.
We know that frequency of simple pendulum follows that 
Now, the speed of the father will be
while for the child the speed will be 
The ratio of the father’s speed to the child’s speed will be

Answer: SG = 2.67
Specific gravity of the sand is 2.67
Explanation:
Specific gravity = density of material/density of water
Given;
Mass of sand m = 100g
Volume of sand = volume of water displaced
Vs = 537.5cm^3 - 500 cm^3
Vs = 37.5cm^3
Density of sand = m/Vs = 100g/37.5 cm^3
Ds = 2.67g/cm^3
Density of water Dw = 1.00 g/cm^3
Therefore, the specific gravity of sand is
SG = Ds/Dw
SG = (2.67g/cm^3)/(1.00g/cm^3)
SG = 2.67
Specific gravity of the sand is 2.67
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The power created is 
Explanation:
From the question we are told that
The that the average power is mathematically represented as

Where W is is the Workdone which is mathematically represented as

Where F is the applies force and s is the displacement due to the force
So

Now this displacement can be represented mathematically as

Where
is the average velocity and
is the time taken
So

=> 
Answer:

Explanation:
Aceleration is a change on the velocity of the object in a given time.
For this case: the initial velocity is

and the final velocity is :

so, the change in velocity is:

and the change in time , according to the problem:

So, the aceleration is:

Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g