Answer:
3m/s
Explanation:
Time=5s
Distance =15m
Speed=distance/time
Putting the values
Speed=15m/5s
Speed=3m/s is the answer
Hope it will help you. :)
Answer:
Thrust = 200 N
Explanation:
The engine thrust can be found by using the following formula:
where,
m = mass flow rate of the fuel = 0.05 kg/s
v = velocity of ejected gases = 4000 m/s
Therefore, using the given values in the equation, we get:
<u>Thrust = 200 N</u>
Answer:
in pounds if would be 50.7 or 50.7063
W=350 J=M g H
Solve for the height, H
G=9.8 m/s^2
M=7kg
H will be in meters if you use those units.
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
<u>ΔS = 49.13 ft</u>
