Answer:
304.89m
Explanation:
Given
acceleration a = 2.52m/s²
final speed v = 39.2m/s
initial speed = 0m/s (car accelerates from rest)
Using the equation of motion below to get the distance of Doc brown from Marty;
v² = u²+2as
substitute the given parameters
39.2² = 0²+2(2.52)s
1536.64 = 0+5.04s
divide both sides by 5.04
1536.64/5.04 = 5.04s/5.04
rearrange the equation
5.04s/5.04 = 1536.64/5.04
s = 304.89m
Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?
Answer with Explanation:
We are given that
Diameter of fighter plane=2.3 m
Radius=
a.We have to find the angular velocity in radians per second if it spins=1200 rev/min
Frequency=
1 minute=60 seconds
Angular velocity=
Angular velocity=
b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

c.Centripetal acceleration=
Centripetal acceleration==
Answer:
Heat of vaporization will be 22.59 j
Explanation:
We have given mass m = 10 gram
And heat of vaporization L = 2.259 J/gram
We have to find the heat required to vaporize 10 gram mass
We know that heat of vaporization is given by
, here m is mass and L is latent heat of vaporization.
So heat of vaporization Q will be = 10×2.259 = 22.59 J