Answer:
1 rev = 2(pi) rad pi(rad) = 180 degrees
so 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s = 63.36 degrees
Explanation:
63.36 estimated to 63 so 63
The correct answer is
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force per unit charge.
In fact, the electric field strength is defined as the electric force per unit charge experienced by a positive test charge located in the electric field. In formula:
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where
E is the electric field strength
F is the electric force experienced by the charge
q is the positive test charge.
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:
W_apparent = 93.1 kg
Explanation:
The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.
W_apparent = W - B
The push is given by the expression of Archimeas
B = ρ_fluide g V
ρ_al = m / V
m = ρ_al V
we substitute
W_apparent = ρ_al V g - ρ_fluide g V
W_apparent = g V (ρ_al - ρ_fluide)
we calculate
W_apparent = 980 50 (2.7 - 0.8)
W_apparent = 93100 g
W_apparent = 93.1 kg
