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sleet_krkn [62]
3 years ago
15

Please give me answer of this question and I will mark you as brainliest ​

Physics
1 answer:
Degger [83]3 years ago
6 0

Answer:

Maharashtra - mashru or himroo / dhoti and lugda

Gujarat - patola / ghagra choli

Punjab - pat / kurta and pajama

Odisha - ikat / Sadi

West Bengal - tossa / kurta

Karnataka - Mysore silk / mundu

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Define a dipole. Hence, write the expression for calculating the electric moment of a dipole​
Aleksandr-060686 [28]

Answer:

Explanation:

An electric dipole is formed by two point charges +q and −q connected by a vector a. The electric dipole moment is defined as p = qa

4 0
2 years ago
g a horizontal wheel of radius is rotating about a vertical axis. What is the magnitude of the resultant acceleration of a bug t
mihalych1998 [28]

Answer:

   a = w² r

Explanation:

In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.

           a = v² / r

angular and linear variables are related

           v = w r

we substitute

          a = w² r

where r is the radius of the wheel

4 0
3 years ago
A car travels up a hill at a constant speed of 38 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the a
mojhsa [17]

Answer:

Average speed will be 48.23 km/h

Explanation:

Let the distance up to hill is = d km

Speed when car goes to hill = 38 km/h

So time required t=\frac{distance}{speed}=\frac{d}{38}hour

Speed when car return from hill = 66 km/h

So time required to return fro hill t=\frac{d}{66}h

Total time t_{total}=\frac{t}{38}+\frac{t}{66}

Total distance = d+d =2d

So average speed=\frac{total\ distance}{total\ time}=\frac{2d}{\frac{d}{38}+\frac{d}{66}}=48.23km/h

8 0
3 years ago
Which of the following are features of the nucleus of an atom
vichka [17]
We r made of atom so v can’t touch anything hehe I just joking
4 0
2 years ago
A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
svp [43]

a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

K=\frac{1}{2}mv^2

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

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

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






5 0
3 years ago
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