Answer:
Well it would be equal to 500N because pushing forward the ball (or whatever maybe a body) would push the canon back an even 500N backwards...
Explanation:
Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.
At the top of the first hill, the car's potential energy is
PE = (mass) x (gravity) x (height) .
At the bottom, the car's kinetic energy is
KE = (1/2) (mass) (speed²) .
You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:
KE = 0.9 PE
(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)
Divide each side by (mass):
(0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)
(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)
Divide each side by (0.9):
(0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)
Divide each side by (9.8 m/s²):
Height = (5/9)(4900 m²/s²) / (9.8 m/s²)
= (5 x 4900 m²/s²) / (9 x 9.8 m/s²)
= (24,500 / 88.2) (m²/s²) / (m/s²)
= 277-7/9 meters
(about 911 feet)
They has been very successful but they are very expensive to operate that is your answer I hope this helps
Answer:
When they are connected in series
The 50 W bulb glow more than the 100 W bulb
Explanation:
From the question we are told that
The power rating of the first bulb is ![P_1 = 100 \ W](https://tex.z-dn.net/?f=P_1%20%20%3D%20100%20%5C%20W)
The power rating of the second bulb is ![P_2 = 50 \ W](https://tex.z-dn.net/?f=P_2%20%20%3D%20%2050%20%5C%20W)
Generally the power rating of the first bulb is mathematically represented as
![P_1 = V^2 R](https://tex.z-dn.net/?f=P_1%20%20%3D%20%20V%5E2%20R)
Where
is the normal household voltage which is constant for both bulbs
So
![R_1 = \frac{V^2}{P_1 }](https://tex.z-dn.net/?f=R_1%20%20%3D%20%20%5Cfrac%7BV%5E2%7D%7BP_1%20%7D)
substituting values
![R_1 = \frac{V^2}{100}](https://tex.z-dn.net/?f=R_1%20%20%3D%20%20%5Cfrac%7BV%5E2%7D%7B100%7D)
Thus the resistance of the second bulb would be evaluated as
![R_2 = \frac{V^2}{50}](https://tex.z-dn.net/?f=R_2%20%20%3D%20%20%5Cfrac%7BV%5E2%7D%7B50%7D)
From the above calculation we see that
![R_2 > R_1](https://tex.z-dn.net/?f=R_2%20%20%3E%20%20R_1)
This power rating of the first bulb can also be represented mathematically as
![P_ 1 = I^2_1 R_1](https://tex.z-dn.net/?f=P_%20%201%20%20%3D%20%20I%5E2_1%20%20R_1)
This power rating of the first bulb can also be represented mathematically as
![P_ 2 = I^2_2 R_2](https://tex.z-dn.net/?f=P_%20%202%20%20%3D%20%20I%5E2_2%20R_2)
Now given that they are connected in series which implies that the same current flow through them so
![I_1^2 = I_2^2](https://tex.z-dn.net/?f=I_1%5E2%20%3D%20%20I_2%5E2)
This means that
![P \ \alpha \ R](https://tex.z-dn.net/?f=P%20%5C%20%5Calpha%20%20%5C%20%20R)
So when they are connected in series
![P_2 > P_1](https://tex.z-dn.net/?f=P_2%20%20%3E%20%20P_1)
This means that the 50 W bulb glows more than the 100 \ W bulb
Answer: Really
Explanation:
Just look it up for this page and maybe you will find an anwser sheet.