To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

Where,
T = Tension
Linear density
Our data are given by
Tension , T = 70 N
Linear density , 
Amplitude , A = 7 cm = 0.07 m
Period , t = 0.35 s
Replacing our values,



Speed can also be expressed as

Re-arrange to find \lambda

Where,
f = Frequency,
Which is also described in function of the Period as,



Therefore replacing to find 


Therefore the wavelength of the waves created in the string is 3.49m
The acceleration due to gravity of Mars is 
<u>Explanation:</u>
As per universal law of gravity, the gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance between them. But in the present case, the gravity need to be determined between Mars and the object on Mars. Since the mass of Mars is greater than the mass of any object. Thus,

Here, G is the gravitational constant, R is the radius of Mars and M, m is the mass of Mars and the object respectively..
Also, according to Newton’s second law of motion, the acceleration of any object will be equal to the ratio of force exerted on it to the mass of the object.
So in order to determine the acceleration due to gravity of Mars, divide the gravitational force of Mars by mass of object on the surface of Mars.




Answer:
So the answer would be water based on the evidence shown below.
Explanation:
Mercury is a poor conductor of heat but good for electricity, water is a good conductor of heat but a poor conductor of electricity, wood is a poor conductor of heat and electricity, and glass is probably the worst conductor of heat.
Answer:
91.87 m/s
Explanation:
<u>Given:</u>
- x = initial distance of the electron from the proton = 6 cm = 0.06 m
- y = initial distance of the electron from the proton = 3 cm = 0.03 m
- u = initial velocity of the electron = 0 m/s
<u>Assume:</u>
- m = mass of an electron =

- v = final velocity of the electron
- e = magnitude of charge on an electron =

- p = magnitude of charge on a proton =

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.
Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.


Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.
The hypotenuse is measured at 120 meters of string, and you need to solve for the leg of the triangle that is horizontal. The degree is 40, so use trigonometry to figure it out.
Cosin (40) is equal to around .766
Adjacent/Hypotenuse
x/120 = cos40
Answer: 91.92533.
If you use 3 significant figures it should be 91.9 meters.