Answer:
A baseball (m= 149g) approaches a bat horizontally at a speed of 40.2 m/s (90 mi/h) and is hit straight back at a speed of 45.6m/s (102mi/h). If the ball is in contact with the bat for a time of 1.10ms, what is the average force exerted on the ball by the bat ? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.
Explanation:
Use the impulse equation (a form of Newton's 2nd Law): FΔt = Δ(mv) where Δ means "change in"
The change in momentum is mBB(vf - vi) = (.150 kg)(-46.9 m/s - 40.5 m/s)
Divide this by the time interval and you get F exerted by the bat in Newtons.
Take care.
I think its [B]
Personally i would say [B] only because If you are looking beyond the car in front of you..... then what if the car in front of you throws on breaks... you would hit them in the butt because you weren't paying attention to the car.
And majority of the time if your looking in the lanes beside you then you are most likely trying to get in that lane.
Answer:
Hey mate......
Explanation:
This is ur answer.....
<h2><em>A. Rotation of Earth</em></h2>
<em>The moon rises in the east and sets in the west, each and every day. It has to. The rising and setting of all celestial objects is due to Earth's continuous daily spin beneath the sky</em><em>.</em>
Hope it helps!
Brainliest pls!
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Answer:
a) - 72.5°c
b) pressure = 3625.13 Pa
c) density = 0.063 kg/m^3
d) it is a subsonic aircraft
Explanation:
a) Determine Temperature
Temperature at 19.5 km ( 19500 m )
T = -131 + ( 0.003 * altitude in meters )
= -131 + ( 0.003 * 19500 ) = - 72.5°c
b) Determine pressure and density at 19.5 km altitude
Given :
Po (atmospheric pressure at sea level ) = 101kpa
R ( gas constant of air ) = 0.287 KJ/Kgk
T = -72.5°c ≈ 200.5 k
pressure = 3625.13 Pa
hence density = 0.063 kg/m^3
attached below is the remaining part of the solution
C) determine if the aircraft is subsonic or super sonic
Velocity ( v ) = 
= 
= 283.8 m/s
hence it is a subsonic aircraft
