Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula ![E = \dfrac{kq }{d}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7Bkq%20%7D%7Bd%7D)
![E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7B9%20%5Ctimes%2010%5E9%20%5Ctimes%20%2833.3%20%5Ctimes%2010%5E%7B-9%7D%29%20%7D%7B0.205%7D)
E = 1461.95 N/C
c) The electric field E is calculated as:
![E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7B9%20%5Ctimes%2010%5E9%20%5Ctimes%20%2833.3%20%5Ctimes%2010%5E%7B-9%7D%29%20%7D%7B1.25%7D)
E = 239.76 N/C
Answer:
<em>The work done by the car is 363 kJ</em>
Explanation:
Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).
Mathematically work done can be expressed as,
E = W = 1/2mv²
W = 1/2mv²................................ Equation 1
Where E = Energy, W = work done, m = mass of the car, v = velocity of the car
<em>Given: m=1500 kg, v=22 m/s</em>
<em>Substituting these values into equation 1</em>
<em>W = 1/2(1500)(22)²</em>
<em>W = 750 × 484</em>
<em>W = 363000 J</em>
<em>W = 363 kJ</em>
<em>Thus the work done by the car is 363 kJ</em>
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
1.549×10-19lJ is the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.
The equation E= hcE =hc, where h is Planck's constant and c is the speed of light, describes the inverse relationship between a photon's energy (E) and the wavelength of light ().
The Rydberg formula is used to determine the energy change.
Rydberg's original formula used wavelengths, but we may rewrite it using units of energy instead. The result is the following.
aaΔE=R(1n2f−1n2i) aa
were
2.17810-18lJ is the Rydberg constant.
The initial and ultimate energy levels are ni and nf.
As a change of pace from
n=5 to n=3 gives us
ΔE
=2.178×10-18lJ (132−152)
=2.178×10-18lJ (19−125)
=2.178×10-18lJ×25 - 9/25×9
=2.178×10-18lJ×16/225
=1.549×10-19lJ
Learn more about Rydberg formula here-
brainly.com/question/13185515
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A) red light
red lights are an example of an electromagnetic wave. visible lights are the only electromagnetic waves we can actually see on the spectrum. red, in particular has the biggest wavelength.
b) ocean waves
ocean waves are not an electromagnetic wave. in fact, it’s a mechanical wave. electromagnetic waves can travel through a vacuum, that is empty space, but mechanical waves cannot.
c) sound waves
sound waves are also not an electromagnetic wave. it’s a mechanical wave. you cannot hear electromagnetic waves.
d) earthquakes
an earthquake is also not an example of electromagnetic waves. it’s a mechanical wave.
hope this helps!