40 meters times 1 meter over 100 centimeters equals 0.4 meters. 1.3 meters + 40 centimeters =. 1.3 m + 0.4 m = 1.7 m. The answer is 1.7 meters
Answer:
18.7842493212 W
Explanation:
T = Tension = 1871 N
= Linear density = 3.9 g/m
y = Amplitude = 3.1 mm
= Angular frequency = 1203 rad/s
Average rate of energy transfer is given by
![P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7BT%5Cmu%7D%5Comega%5E2y%5E2%5C%5C%5CRightarrow%20P%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B1871%5Ctimes%203.9%5Ctimes%2010%5E%7B-3%7D%7D%5Ctimes%201203%5E2%5Ctimes%20%283.1%5Ctimes%2010%5E%7B-3%7D%29%5E2%5C%5C%5CRightarrow%20P%3D18.7842493212%5C%20W)
The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W
Using Ampere's Law, the magnetic field produced inside this solenoid is given by
B = uo N I / h
where uo is the vacuum permeability, N is the number of turns in the solenoid and h is the length of the solenoid. Earth's magnetic field is around 50 microteslas in North America thus the current needed in the solenoid is
I = B h / (uo N) = (50 E-6 ) (4) / ((4 pi E-7)(6000) ) = 0.026 A
I = 26 mA
So you need a current of around 26 mA.
Answer:
high, low
Explanation:
- Energy always flows from a higher level to a lower level.
- It is analogous to the waterfall where waterfalls from a higher level to a lower level.
- So in the case of the pressure of the gas, when there are any numbers of molecules in a given volume of space. The gas is said to be at high pressure.
- When there are fewer molecules in the given volume. The gas is said to be at lower pressure.
- Due to a large number of atoms, the high-pressure gas exerts more force on the container than the force exerted by the low-pressure gas.
- If a hose is connected between these two containers, gas rushes from high pressure to the low pressure. Since the force exerted by the high-pressure gas is greater than that of low-pressure gas.
So, the wind tends to move from high-pressure areas to low pressure.
To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,
![\mu = \frac{f_0}{f_e}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7Bf_0%7D%7Bf_e%7D)
Here,
= Magnification
= Focal length eyepieces
= Focal length of the Objective
Rearranging to find the focal length of the objective
![f_0 = \mu f_e](https://tex.z-dn.net/?f=f_0%20%3D%20%5Cmu%20f_e)
Replacing with our values
![f_0 = 7.5* 3.7cm](https://tex.z-dn.net/?f=f_0%20%3D%207.5%2A%203.7cm)
![f_0 = 27.75cm](https://tex.z-dn.net/?f=f_0%20%3D%2027.75cm)
Therefore the focal length of th eobjective lenses is 27.75cm