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3 years ago

I NEED HELP RIGHT NOW PLEASE I will 30pts and A brainly to anyone who helps me with questions 2-13!!!!!!!

Physics

2 answers:

neonofarm [45]3 years ago

3 0

True True False True False False True I hope I helped on the first few

Evgesh-ka [11]3 years ago

3 0

w o ao

hope this helps

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A gram is the basic metric unit for measuring which of the following?

DanielleElmas [232]

Gram is used for measuring weight

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3 years ago

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A copper wire is 1.6 m long and its diameter is 1.1 mm. If the wire hangs vertically, how much weight (in N) must be added to it

qaws [65]

Answer:

Weight required = 194.51 N

Explanation:

The elongation is given by

$\Delta L=\frac{PL}{AE}$

Length , L= 1.6 m

Diameter, d = 1.1 mm

Area

$A=\frac{\pi d^2}{4}=\frac{\pi \times (1.1\times 10^{-3})^2}{4}=9.50\times 10^{-7}m^2$

Change in length, ΔL = 2.8 mm = 0.0028 m

Young's modulus of copper, E = 117 GPa = 117 x 10⁹ Pa

Substituting,

$\Delta L=\frac{PL}{AE}\\\\0.0028=\frac{P\times 1.6}{9.50\times 10^{-7}\times 117\times 10^9}\\\\P=194.51N$

Weight required = 194.51 N

8 0

3 years ago

The oscillating electric field in a plane electromagnetic wave is given by <img src="https://tex.z-dn.net/?f=%7B50%5Csin%20%28%5

Scorpion4ik [409]

Here, we are given with:

${:\implies \quad \longrightarrow \begin{cases}\sf E_{0}= 50\: V/m\\ \sf \nu = 2\times 10^{7}Hz\end{cases}}$

(a) So now, we can thus obtain

${:\implies \quad \sf \omega =2\pi \nu}$

${:\implies \quad \sf \omega =2\pi (2\times 10^{7})}$

${:\implies \quad \boxed{\bf{\omega = 4\pi \times 10^{7}\: s^{-1}}}}$

Now, finding $\lambda$

${:\implies \quad \sf \lambda =\dfrac{c}{\omega}=\dfrac{3\times 10^{8}}{4\pi \times 10^{7}}}$

${:\implies \quad \boxed{\bf{\lambda \approx 2.39\:m}}}$

Now, finding k

${:\implies \quad \sf k=\dfrac{2\pi}{\lambda}=\dfrac{200\pi}{239}}$

${:\implies \quad \boxed{\bf{k\approx 2.63\:m^{-1}}}}$

Thus, expression for the electric field is:

${:\implies \quad \boxed{\bf{E=50\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)}}}$

(b) Now, here

${:\implies \quad \sf B_{0}=\dfrac{E_{0}}{c}=\dfrac{50}{3\times 10^{8}}}$

${:\implies \quad \boxed{\bf{B_{0}\approx 16.67×10^{-7}\:T}}}$

Thus, expression for the magnetic field:

${:\implies \quad \boxed{\bf{B_{y}=16.67\times 10^{-7}\sin \bigg(4\pi \times 10^{7}t-\dfrac{263x}{100}\bigg)\:T}}}$

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2 years ago

Two parallel circular plates with radius carrying equal-magnitude surface charge densities of are separated by a distance of How

SpyIntel [72]

Answer:

I guess it is A. I am not sure

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3 years ago

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Describe the relationship between the length and period of a pendulum in the language of direct proportions

Wittaler [7]

The period of the pendulum is directly proportional to the square root of the length of the pendulum

Explanation:

The period of a simple pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that when the length of the pendulum increases, the period of the pendulum increases as the square root of L, $T\propto \sqrt{L}$ . This means that

The period of the pendulum is directly proportional to the square root of the length of the pendulum

From the equation, we also notice that the period of a pendulum does not depend on its mass.

#LearnwithBrainly

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3 years ago