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2 years ago

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The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin

e sine 2 pi ft comma for time t in seconds, where i is instantaneous current in amperes, Imax is maximum current in amperes and f is the number of cycles per second. Find the least positive value of t, given the following data. i=180, Imax=200, f=50

1 answer:

KiRa [710]2 years ago

8 0

Answer:

Explanation:

i = Imax sin2πft

given i = 180 , Imax = 200 , f = 50 , t = ?

Put the give values in the equation above

180 = 200 sin 2πft

sin 2πft = .9

sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

360 x 50 t = 360n + 64

360 x 50 t = 64 , ( putting n = 0 for least value of t )

18000 t = 64

t = 3.55 ms .

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A classical estimate of the vibrational frequency is ff = 7.0×10137.0×1013 HzHz. The mass of a hydrogen atom differs little from

vlada-n [284]

Answer:

The force constant of the spring is 317.8 N/m.

Explanation:

Given that,

Frequency $f=7.0\times10^{13}\ Hz$

We need to calculate the reduced mass

Using formula of reduced mass

$\mu=\dfrac{m_{H}m_{I}}{m_{H}+m_{I}}$

Where, $m_{H}$ = atomic mass of H

$m_{I}$ = atomic mass of I

Put the value into the formula

$\mu=\dfrac{1\times126.9}{1+126.9}$

$\mu=0.99\ u$

$\mu=0.99\times1.66\times10^{-27}\ Kg$

$\mu=1.643\times10^{-27}\ kg$

We need to calculate the force constant of the spring

Using formula of frequency

$f=\dfrac{1}{2\pi}\times\sqrt{\dfrac{k}{\mu}}$

$k=f^2\times 4\pi^2\times\mu$

Put the value into the formula

$k=(7.0\times10^{13})^2\times4\pi^2\times1.643\times10^{-27}$

$k=317.8\ N/m$

Hence, The force constant of the spring is 317.8 N/m.

0 0

2 years ago

5. A nail contains trillions of electrons. Given that electrons repel from each other, why do they not then fly out of the nail?

diamong [38]

Answer:

Nails are made of iron. Iron consists of 26 protons and 26 electrons. protons are positively charged and electrons are negatively charged, so this force of attraction keeps the electrons together.

If electrons repel from each other, the positively charge protons and nucleus allow them to move in a definite orbit and prevent them flying out of the nail.

4 0

1 year ago

What is true of an object pulled inward in an electric field?

slava [35]

Answer:

option b

Explanation:

There is an object pulled inward in an electric field.

We have to find out of the four options given which is true.

a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.

b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object

c) The object has a negative charge will be correct only if the original charge was positive hence wrong

d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge

So only option b is right

5 0

2 years ago

7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh

pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

$F= \mu R$

where $\mu$ is the coefficient of friction = 0.02

R = Normal reaction of the load = $mgcos\theta$ = $25 \times 9.81 \times cos 10$ = $241.52N$

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

$F=0.02 \times 241.52N$

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0

2 years ago

A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision

TEA [102]

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

$0 = m(3.5 sin30) - mv_y$

$v_y = 1.75 m/s$

Now similarly in X direction we will have

$m(4.5) = m(3.5 cos30 ) + mv_x$

$v_x = 1.47 m/s$

now final kinetic energy of both puck after collision is given as

$KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)$

$KE_f = 8.73 m$

initial kinetic energy of both pucks is given as

$KE_i = \frac{1}{2}m(4.5^2) + 0$

$KE_i = 10.125 m$

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

5 0

2 years ago