Answer:
the distance that the object is raised above its initial position is 5.625 m.
Explanation:
Given;
applied effort, E = 15 N
load lifted by the ideal pulley system, L = 16 N
distance moved by the effort, d₁ = 6 m
let the distance moved by the object = d₂
For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.
M.A = V.R

Therefore, the distance that the object is raised above its initial position is 5.625 m.
Answer:
Answer: The spring constant of the spring is k = 800 N/m, and the potential energy is U = 196 J. To find the distance, rearrange the equation: The equation to find the distance the spring has been compressed is therefore: The spring has been compressed 0.70 m, which resulted in an elastic potential energy of U = 196 J being stored.
Explanation:
The answer is B) region of high pressure in a medium caused by a passing wave
Compression is the forcing of the molecules of a medium, be it water, air, or something else, as a wave passes by. This forcing together of the molecules raises the pressure of the medium in the area that the wave passes through.
The velocity when function p(t)=11 is 8 .
According to the question
The position of a car at time t represented by function :
Now,
When function p(t) = 11 , t will be
11 = t²+2t-4
0 = t² + 2t - 15
or
t² +2t-15 = 0
t² +(5-3)t-15 = 0
t² +5t-3t-15 = 0
t(t+5)-3(t+5) = 0
(t-3)(t+5) = 0
t = 3 , -5
as t cannot be -ve as given ( t≥0)
so,
t = 3
Now,
the velocity when p(t)=11
As we know velocity =
therefore to get the value of velocity from function p(t)
we have to differentiate the function with respect to time
v(t) = 2t + 2
where v(t) = velocity at that time
as t = 3 for p(t)=11
so ,
v(t) = 2t + 2
v(t) = 2*3 + 2
v(t) = 8
Hence, the velocity when function p(t)=11 is 8 .
To know more about function here:
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Answer:
q₃=5.3nC
Explanation:
First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

In words, the value of q₃ must be 5.3nC.