Answer:
a) The student feel light
b) Nbottom = 758 N
c) N'top= 236 N
d) N'bottom= 1055 N
Explanation:
a) W= 659N , Ntop= 560N
W > Ntop ---> Student feel less weight
b) Top:
∑F= W - Ntop = m.v²/R
m.v²/R = 659N - 560 N = 99 N
Bottom:
∑F= Nbottom- W = m.v²/R
Nbottom= W + m.v²/R = 659N + 99 N = 758N
c) W= 659 N , Ntop= 560 N , v'=2.v
N'top= ?
∑F= W - N'top = m.v'²/R
N'top= W - 4.m.v²/R
N'top = 659 N - 4. 99 N = 263 N
d) N'bottom = ?
∑Fbottom= N'bottom- W = m.v'²/R
N'bottom = W + 4.m.v²/R = 659 N + 4. 99 N = 1055 N
Answer:
The charge is 
Explanation:
Given that,
Distance = 2.5 mm
Electric field = 800 NC
Length 
We need to calculate the linear charge density
Using formula of linear charge density
Put the value into the formula
We need to calculate the charge
Using formula of charge
Put the value into the formula
Hence, The charge is
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>
Answer:
Airplane wings must be designed to ensure that air molecules move more rapidly over the top surface of the wing, creating a region of lower pressure.
Explanation:
Answer:
0.707m
Explanation:
from formula of range i.e R=Usin2Q/g
