Answer:
d = 19.92 m
Explanation:
As in this exercise there is friction we must use the relationship between work and energy
W = ΔEm
Look for energy in two points
Initial. Fully compressed spring
Em₀ = 
= ½ k x²
Final. When the block stopped
= 0
Let's look for the work of the rubbing force
W = fr d cos θ
Since rubbing is always contrary to movement, θ = 180
W = - fr d
Let's use Newton's second Law, to find the force of friction
Y Axis
N- w = 0
N = mg
The equation for the force of friction is
fr = μ N
fr = μ mg
We substitute in the work equation
W = - μ m g d
We write the relationship of work and energy
-μ m g d = 0 - ½ k x²
d = ½ k x² / μ m g
Let's calculate
d = ½ 131 2.1 2 / (0.74 2 9.8)
d = 19.92 m
Answer: 1 000 times
Explanation: 1 000 centimetre is equal to 1 dekacemetre.
Using conversion will apply in converting every number.
To have a force of 3.8x10-8 the two chaarges must be 0.65m apart.
Answer:
343/1500
Explanation:
Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).
From the question,
P' = mg×v................. Equation 1
Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.
Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²
Substitute into equation 1
P' = 700(2.5)(9.8)
P' = 17150 W.
If the full power generated by the engine = 75000 W
The fraction of the engine power used to make the climb = 17150/75000
= 343/1500
The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,
d = Vot + 0.5at²
where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,
6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²
Determining the value of a,
a = 1.646 m/s²
ANSWER: 1.646 m/s²
