Answer:
W=76.55 miles.metric tons
Explanation:
Given that
Weight on the earth = 12 tons
So weight on the moon =12/6 = 2 tons
( because at moon g will become g/6)
As we know that

Here x= 1100 miles
F 2 tons

So

We know that
Work = F. dx


![W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7Bx%7D%5Cright%5D_%7B1100%7D%5E%7B1140%7D)
![W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7B1140%7D-%5Cdfrac%7B1%7D%7B1100%7D%5Cright%5D)
W=76.55 miles.metric tons
Answer:
Doubling the voltage in this arrangement both doubles the voltage drop across the resistor and the current through it. The bulb will be much brighter.
Answer:
The magnitude of the resistive force exerted by the water is
newtons.
Explanation:
By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:
(Eq. 1)
Where:
- Forward trust, measured in newtons.
- Resistive force exerted by the water, measured in newtons.
From (Eq. 1), we get that: (
)


The magnitude of the resistive force exerted by the water is
newtons.
I believe it’s force but i’m not really sure.
Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)

where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:

0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:

Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:

5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:

10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:

10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]