Answer:
600m
Explanation:
30×20 at a constant speed is 600m.
Answer:
B. QC > 0; QH < 0
Explanation:
Given that there are two reservoir of energy.
Sign convention for heat and work :
1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.
2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.
From hot reservoir heat is going out that is why it is taken as negative
![Q_H](https://tex.z-dn.net/?f=Q_H%3C0)
From cold reservoir heat is coming inside the reservoir that is why it is taken as positive
![Q_C>0](https://tex.z-dn.net/?f=Q_C%3E0)
That is why the answer will be
,![Q_C>0](https://tex.z-dn.net/?f=Q_C%3E0)
To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.
If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.
![F = ma \rightarrow a = \frac{F}{m}](https://tex.z-dn.net/?f=F%20%3D%20ma%20%5Crightarrow%20a%20%3D%20%5Cfrac%7BF%7D%7Bm%7D)
Replacing,
![a =\frac{42N}{83000kg}](https://tex.z-dn.net/?f=a%20%3D%5Cfrac%7B42N%7D%7B83000kg%7D)
![a =5.06*10^{-4}m/s^2](https://tex.z-dn.net/?f=a%20%3D5.06%2A10%5E%7B-4%7Dm%2Fs%5E2)
The total speed change
we have that the value is 0.71m/s
If we know that acceleration is the change of speed in a fraction of time,
![a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}](https://tex.z-dn.net/?f=a%3D%20%5Cfrac%7B%5CDelta%20v%7D%7Bt%7D%20%5Crightarrow%20t%20%3D%20%5Cfrac%7B%5CDelta%20v%7D%7Ba%7D)
We have that,
![t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7B0.71m%2Fs%7D%7B5.06%2A10%5E%7B-4%7Dm%2Fs%5E2%20%7D)
![t = 1403.16s](https://tex.z-dn.net/?f=t%20%3D%201403.16s)
Therefore the Rocket should be fired around to 1403.16s
Answer:
a. 120 W
b. 28.8 N
Explanation:
To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance. Suppose a cyclist is traveling at 15 km/h on level ground. Assume he is using 480 W of metabolic power.
a. Estimate the amount of power he uses for forward motion.
b. How much force must he exert to overcome the force of air resistance?
(a)
He is 25% efficient, therefore the cyclist will be expending 25% of his power to drive the bicycle forward
Power = efficiency X metabolic power
= 0.25 X 480
= 120 W
(b)
power if force times the velocity
P = Fv
convert 15 km/h to m/s
v = 15 kmph = 4.166 m/s
F = P/v
= 120/4.166
= 28.8 N
definition of terms
power is the rate at which work is done
force is that which changes a body's state of rest or uniform motion in a straight line
velocity is the change in displacement per unit time.
Answer:
B. space quantization.
Explanation:
In 1921, Otto Stern developed the idea behind this experiment, while Walther Gerlach performed the actual experiment in 1922. The Ster-Gerlach experiment provides prove to the fact that the spatial orientation of angular momentum is quantized. To demonstrate the experiment, silver atoms were made to travel through a magnetic field path.
Before they hit the screen(usually a glass slide), they were deflected because of their non-zero magnetic moment. There was an expected result for this experiment, but the actual observation on the glass slide was a continuous distribution of the silver atoms that actually hit the glass. This experiment was useful in proving that in all atomic-scale systems, there was a quantization of angular momentum.