Answer:
160 kg
12 m/s
Explanation:
= Mass of first car = 120 kg
= Mass of second car
= Initial Velocity of first car = 14 m/s
= Initial Velocity of second car = 0 m/s
= Final Velocity of first car = -2 m/s
= Final Velocity of second car
For perfectly elastic collision
Applying in the next equation
Mass of second car = 160 kg
Velocity of second car = 12 m/s
Answer:
50 meters
Explanation:
Let's start by converting to m/s. There are 3600 seconds in an hour and 1000 meters in a kilometer, meaning that 72km/h is 20m/s.
Since the car starts at rest, you can write the following equation:
Now that you have the acceleration, you can do this:
Once again, there is no initial velocity:
Hope this helps!
Answer:
The required new pressure is 775 mm hg.
Explanation:
We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.
We have to find the required new pressure.
Let the required new pressure be ''.
As we know that Boyle's law formula states that;
where, = original pressure of gas in the container = 310 mm hg
= required new pressure
= volume of gas in the container = 185 ml
= desired new volume of the gas = 74 ml
So,
= 775 mm hg
Hence, the required new pressure is 775 mm hg.
Answer:
so the transverse displacement is 0.089963 m
Explanation:
Given data
equation y(x,t)=Acos(kx−ωt)
speed v = 9.00 m/s
amplitude A = 9.00 × 10^−2 m
wavelength λ = 0.480 m
to find out
the transverse displacement
solution
we know
v = angular frequency / wave number
and
wave number = 2 / λ = 2
/ 0.480 = 13.0899
angular frequency = v k
angular frequency = 9.00 × 13.0899
angular frequency = 117.8097 rad/sec = 118 rad/sec
so
equation y(x,t)=Acos(kx−ωt)
y(x,t)=9.00 × 10^−2 cos(13.0899 x−118t)
when x =0 and and t = 0
maximum y(x,t)= 9.00 × 10^−2 cos(13.0899 (0) − 118 (0))
maximum y(x,t)= 9.00 × 10^−2 m
and when x = x = 1.59 m and t = 0.150 s
y(x,t)=9.00 × 10^−2 cos(13.0899 (1.59) −118(0.150) )
y(x,t)=9.00 × 10^−2 × (0.99959)
y(x,t) = 0.089963 m
so the transverse displacement is 0.089963 m