In order to find the radius of the coin, we need to use:
Ac = V^2 r
In which,
Ac = acceleration of the coin = 2.2 m/s^2
V= rotational Speed = (18/12) * 2πr
r= Radius
so,
<span>2.2 = 9(π^2)(r^2) / r
</span><span>= 9(π^2)r
</span>
<span>r = 2.2 / 9(π^2) = 0.02476740044 m . . .or you can rounded it up to 0.025 m</span>
use Newton's gravitational law and 2nd law .....
Momentum = (mass) x (velocity)
Original momentum before the hit =
(0.16 kg) x (38 m/s) this way <==
= 6.08 kg-m/s this way <==
Momentum after the hit =
(0.16) x (44 m/s) that way ==>
= 7.04 kg-m/s that way ==>
Change in momentum = (6.08 + 7.04) = 13.12 kg-m/s that way ==> .-----------------------------------------------
Change in momentum = impulse.
Impulse = (force) x (time the force lasted)
13.12 kg-m/s = (force) x (0.002 sec)
(13.12 kg-m/s) / (0.002 sec) = Force
6,560 kg-m/s² = 6,560 Newtons = Force
( about 1,475 pounds ! ! ! )
Hoped this helped!! ☺
If E = 1/2 * m * v^2
v = (2E/m)^1/2
so the larger the mass, the higher the velocity hence taylor is moving faster
Answer:
Yes, the race car driver needs a faster reaction time than someone driving in a school zone.
Explanation.
For the sake of argument, let us consider
(i) a person driving at 35 mph in a school zone (as a normal driver);
(ii) a person driving at 60 mph in a school zone (as a racing driver).
Suppose a blind pedestrian crosses the road 0.1 miles (about 500 feet) in front of the driver.
The time before the normal driver hits the pedestrian is
(0.1 /35)*3600 = 10.3 seconds.
The time before the racing driver hits the pedestrian is
(0.1/60)*3600 = 6 seconds.
Because a reaction time of 6 seconds may be insufficient to avoid hitting the pedestrian, the racing driver needs a faster reaction time than the normal driver.
