Answer:
T_finalmix = 59.5 [°C].
Explanation:
In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.
where:
mwater = mass of the water = 0.4 [kg]
Cp_water = specific heat of the water = 4180 [J/kg*°C]
T_waterinitial = initial temperature of the water = 85 [°C]
T_finalmix = final temperature of the mix [°C]
Now replacing:
![0.4*4180*(85-T_{final})=0.4*2450*(T_{final}-16)\\142120-1672*T_{final}=980*T_{final}-15680\\157800=2652*T_{final}\\T_{final}=59.5[C]](https://tex.z-dn.net/?f=0.4%2A4180%2A%2885-T_%7Bfinal%7D%29%3D0.4%2A2450%2A%28T_%7Bfinal%7D-16%29%5C%5C142120-1672%2AT_%7Bfinal%7D%3D980%2AT_%7Bfinal%7D-15680%5C%5C157800%3D2652%2AT_%7Bfinal%7D%5C%5CT_%7Bfinal%7D%3D59.5%5BC%5D)
Answer:
d A ball is rolling down an inclined plane.
Explanation:
When path length is equal to the displacement
then we can say that the motion of the object must be in straight line so that the distance and displacement must be same
SO here we can say
a A ball on the end of a string is moving in a vertical circle.
In circular path distance and displacement is not same
b A toy train is traveling around a circular track.
In circular path distance and displacement is not same
c A train travels 5 miles east before it stops. It then travels 2 miles west.
Net displacement is 3 miles East while distance is 7 miles
d A ball is rolling down an inclined plane.
Here its motion is in straight line so we can say that path length and displacement will be same
e A ball rises and falls after being thrown straight up from the earth's surface.
In this type of to and fro motion path length is not same as displacement
The relationship between the distance covered, initial and final speeds, and time can be expressed through the equation,
First equation,
2ad = Vf² - Vi²
Substituting the known values,
2(a)(0.230 km) = (70 km/h)² - (40 km/h)²
The value of a from the equation is 7173.92 km/h².
Second equation,
d = (Vi)(t) + 0.5at²
Substituting the known values,
0.230 km = (40 km/h)(t) + (0.5)(7173.92 km/h²)(t²)
The value of t from the equation is 4.1818 x 10^-3 hours which is also equal to 0.2509 minutes or 15 seconds.
Answer: 15 seconds
Answer:
ΔE = 37.8 x 10^9 J
Explanation:
The energy required will increased the potential energy and increase the kinetic energy.
As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative
Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²
G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²
ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J
The centripetal force at orbit must be equal to the gravity force
mv²/R = mg'
v²/8.0e6 = 6.272
v² = (6.272(8.0e6)) = 50.2e6 m²/s²
The maximum velocity when resting on earth at the equator is about 460 m/s.
The change in kinetic energy is
ΔKE = ½m(vf² - vi²)(1000)
ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J
Total energy increase is
25e9 + 12.857e9 = 37.8e9 J
Answer:
D. Checking to see if the brake fluid is contaminated
Explanation:
