The answer is D. The temperature obviously doesnt rise slower or faster, and if you are heating an object, it would make no sense to say that less heat is being transferred.
Answer:
The fireman will continue to descend, but with a constant speed.
Explanation:
In kinetic friction <em>(which is the case discussed here) </em>since the fireman is already in motion because of a certain force, once the frictional force matches the normal force, the fireman will stop accelerating and continue moving at a constant rate with the original speed he had. We will need a force greater than the normal force acting on the fireman to cause a deceleration.
We need to understand the difference between static friction and kinetic friction.
Static friction occurs in objects that are stationary, while kinetic friction occurs in objects that are already in motion.
In static friction, when the frictional force matches the weight or normal force of the object, the object remains stationary.
While in kinetic friction, when the frictional force matches the normal force, the object will stop accelerating. This is the case of the fireman sliding down the pole as discussed above.
The yo-yo speeds up when you rub it
Answer:
0.16 m
Explanation:
A rectangular gasoline tank can hold 50.0 kg of gasoline when full, and the density of gasoline is 6.8 × 10² kg/m³. We can find the volume occupied by the gasoline (volume of the tank).
50.0 kg × (1 m³/6.8 × 10² kg) = 0.074 m³
The volume of the rectangular tank is:
volume = width × length × depth
depth = volume / width × length
depth = 0.074 m³ / 0.500 m × 0.900 m
depth = 0.16 m
Answer:
Explanation:
The speed of light in these mediums shall be lower than that in vacuum thus the total time light needs to cross both the media are calculated as under
Total time = Time taken through ice + Time taken through quartz
Time taken through ice = Thickness of ice / (speed of light in ice)
Thus in the same time the it would had covered a distance of
![Distance_{vaccum}=Totaltime\times V_{vaccum}\\\\Distance_{vaccum}=10^{-2}[2.20\mu _{ice+1.50\mu _{quartz}}]](https://tex.z-dn.net/?f=Distance_%7Bvaccum%7D%3DTotaltime%5Ctimes%20V_%7Bvaccum%7D%5C%5C%5C%5CDistance_%7Bvaccum%7D%3D10%5E%7B-2%7D%5B2.20%5Cmu%20_%7Bice%2B1.50%5Cmu%20_%7Bquartz%7D%7D%5D)
we have
Applying values we have
![Distance_{vaccum}=10^{-2}[2.20\times 1.309+1.50\times 1.542]](https://tex.z-dn.net/?f=Distance_%7Bvaccum%7D%3D10%5E%7B-2%7D%5B2.20%5Ctimes%201.309%2B1.50%5Ctimes%201.542%5D)
