Answer:
after it has hit the ground
Heat lost or gained, H = mc(θ₂ - θ₁)
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature
m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C (Since it was cooled).
H = 6.9 kj = 6.9 *1000J = 6900 J
6900 = 200*0.444* (θ₂ - 22)
6900/(200*0.444) = θ₂ - 22
77.70 = θ₂ - 22
θ₂ - 22 = 77.7
θ₂ = 77.7 + 22 = 99.7
So initial temperature before cooling ≈ 100°C . Option C.
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>
Can i have more information?
Answer:
c)
Explanation:
As we know that resultant force is the net force that is acting on the system
As per Newton's II law we know that net force is product of mass and acceleration
so we will have
here we know
m = 80 kg
for circular motion acceleration is given as
now we have
