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3 years ago

How do electromagnetism and gravitation differ from the strong and weak nuclear forces?

1 answer:

5 0

Electromagnetism is the interaction between charged particles. It is very long ranged. It can manifest as a force causing like charges to repel and unlike charges to attract. ... The weak nuclear force is very short ranged and is not a force at all.

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A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v

Tpy6a [65]

Current will be $I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz$
now just pluf in the values and Voila..

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3 years ago

A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol

PIT_PIT [208]

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 $E = \dfrac{kq }{d}$

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}$

E = 1461.95 N/C

c) The electric field E is calculated as:

$E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}$

E = 239.76 N/C

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3 years ago

aev [14]

OD because Boyle’s law specifically states

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3 years ago

A helicopter has blades of length 4.0 m rotating at 3.0 rev/s in a horizontal plane.If the vertical component of the Earth’s mag

VARVARA [1.3K]

Answer:

Induced emf, $\epsilon=9.79\times 10^7\ volts$

Explanation:

Given that,

Length of the helicopter, l = 4 m

Angular speed of the helicopter, $\omega=3\ rev/s=18.84\ rad/s$

The vertical component of the Earth’s magnetic field is, $B=6.5\times 10^5\ T$

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

$\epsilon=\dfrac{1}{2}B\omega l^2$

$\epsilon=\dfrac{1}{2}\times 6.5\times 10^5\times 18.84\times (4)^2$

$\epsilon=9.79\times 10^7\ volts$

So, the induced emf between the tip of a blade and the hub is $\epsilon=9.79\times 10^7\ volts$ . Hence, this is the required solution.

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3 years ago

Q5- Knowing now that your mood might affect your memory, what strategies could you employ to assist your memories and make sure

Tema [17]

You can try recalling and rehearsing information about the memory.

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2 years ago