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mr Goodwill [35]
3 years ago
5

A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a

line joining the two charges is the electric field zero?
Physics
1 answer:
Maksim231197 [3]3 years ago
3 0

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

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The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o
Leya [2.2K]
Answer:
frequency = 5.52 * 10² Hz

Explanation:
the equation that relates velocity, frequency and wavelength is:
velocity = frequency * wavelength

We are given that:
velocity = 331 m/sec
wavelength = 0.6 m

Substitute with the givens in the equation to get the frequency as follows:
velocity = frequency * wavelength
331 = frequency * 0.6
frequency = 331 / 0.6
frequency = 5.52 * 10² Hz

Hope this helps :)
4 0
3 years ago
An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)
Alecsey [184]
"60 kg" is not a weight.  It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is   

                                 (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

                   Weight = (60 kg) x (9.8 m/s²)

                                =      588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to  5R  from the center, the gravity out there is

                    (1R/5R)²  =  (1/5)²  =  1/25  =  0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

                 (0.04) x (588 Newtons)  =  23.52 Newtons.

Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink. 
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
6 0
3 years ago
1. What is the wavelength of a sound wave with a frequency of 50 Hz, if the Speed of sound is 343 m/s.
marshall27 [118]

1.6.86

2.59.04

3.3

Thats the answer I think

7 0
3 years ago
Acceleration due to gravity is stated as a constant which is given 9.8 m/s2. Many scientists have planned and design experiments
stich3 [128]
The answer is 98.1 m/s2.
3 0
3 years ago
What's is the kinetic energy of a .235-kg baseball thrown at 50.0m/s
san4es73 [151]

Answer:

34.51

Explanation:

k=1/2mv² is the kenetic energy equation to fill is in

k=[1/2(0.235)×50]²

6 0
3 years ago
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