(a) Let's convert the final speed of the car in m/s:

The kinetic energy of the car at t=19 s is

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:

But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into

(c) The instantaneous power is given by

where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:

And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
The tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.
<h3>
What is the tension in the cord?</h3>
The tension in the cord is calculated as follows;
T = ma + mg
where;
- a is the acceleration of the block
- g is acceleration due to gravity
- m is mass of the block
T = m(a + g)
T = 1.5(a + 9.8)
T = 1.5a + 14.7
Thus, the tension in the cord is (1.5a + 14.7) N.
If the block is at rest, the tension is 14.7 N.
<h3>Force of the force</h3>
The force with which the cord pulls is equal to the tension in the cord
F = T = m(a + g)
F = (1.5a + 14.7) N
If the block is stationary, a = 0, the tension and force of pull of the cord = 14.7 N.
Thus, the tension in the cord is 14.7 N and the force of pull of the cord is 14.7 N, assuming the block is stationary.
Learn more about tension here: brainly.com/question/187404
#SPJ1
Answer:
Jesus is in everyone's heart...
Just try to find him..:D
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s