Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
A-200 kg
I hope this helped xx
Answer:
The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is 
ft/ sec
Explanation:
Given:
h(t) = 25 ft/sec
x(t) = 10 ft/ sec
h(5) = 25 ft/sec . 5 = 125 ft
x(5) = 10 ft/sec . 5 = 50 ft
Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem
Lets find the derivative of distance with respect to time
Substituting the values of h(t) and x(t) and simplifying we get,
= 
= ft / sec
Answer:
1456 N
Explanation:
Given that
Frequency of the piano, f = 27.5 Hz
Entire length of the string, l = 2 m
Mass of the piano, m = 400 g
Length of the vibrating section of the string, L = 1.9 m
Tension needed, T = ?
The formula for the tension is represented as
T = 4mL²f²/ l, where
T = tension
m = mass
L = length of vibrating part
F = frequency
l = length of the whole part
If we substitute and apply the values we have Fri. The question, we would have
T = (4 * 0.4 * 1.9² * 27.5²) / 2
T = 4368.1 / 2
T = 1456 N
Thus, we could conclude that the tension needed to tune the string properly is 1456 N
It would be Joules.
Workdone is measured in Joules.
Workdone = Force * distance
Force = mass * acceleration
= kg * ms⁻²
= kgms⁻²
Distance = m
So, Force * distance
kgms⁻² * m
Apply laws of indices that says
x² * x³ = x²⁺³ = x⁵
Therefore, It would be kgm²s⁻²
m¹ * m¹ = m¹⁺¹ = m²
s⁻² is also = s / 2
