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3 years ago

The maximum current output of a 60 ω circuit is 11 A. What is the rms voltage of the circuit?

Physics

1 answer:

bezimeni [28]3 years ago

8 0

Answer:

660V

Explanation:

V=IR

V=11×60

=660V

hope this helps

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You have just landed on Planet X. You take out a ball of mass 101 g , release it from rest from a height of 16.0 m and measure t

IRINA_888 [86]

Answer:

0.3817 N

Explanation:

Remark

One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.

Givens

m = 101 g

vi = 0 That's what at rest means.

t = 2.91 s

d = 16 m

F= ?

Formulas

d = vi*t + 1/2*a * t^2

Force = m * a

Solution

16 = 0 + 1/2 a * 2.91^2

16 = 4.234 a Divide by 4.234

16/4.234 = a

a = 3.779

F = m * a

a = 3.779

m = 101 g = 1 kg / 1000 grams

m = 0.101 kg

F = 0.101 * 3.779

F = 0.3817N

8 0

3 years ago

A wheel that was initially spinning is accelerated at a constant angular acceleration of 5.0 rad/s^2. After 8.0 s, the wheel is

notka56 [123]

Answer:

a) Initial angular speed = 30 rad/s

b) Final angular speed = 70 rad/s

Explanation:

a) We have equation of motion s = ut + 0.5at²

Here s = 400 radians

t = 8 s

a = 5 rad/s²

Substituting

400 = u x 8 + 0.5 x 5 x 8²

u = 30 rad/s

Initial angular speed = 30 rad/s

b) We have equation of motion v = u + at

Here u = 30 rad/s

t = 8 s

a = 5 rad/s²

Substituting

v = 30 + 5 x 8 = 70 rad/s

Final angular speed = 70 rad/s

8 0

3 years ago

Which of the following statements cannot be supported by Kepler's laws of planetary motion? a. The distance of a planet around t

hodyreva [135]

i guess option (a)

is the right one

if it's correct

select me as brainliest

5 0

3 years ago

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

$A = \pi r^2$

The magnetic flux is given by:

$\phi = \int \vec{B} \cdot d\vec{A}$

$d\vec{A}$ is parallel to $\vec{B}$ and $\vec{B}$ is constant in magnitude and direction therefore:

$\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2$

Part A)

initially the flux is $\phi =\pi B r^2$

after the interval $\Delta t= 2.4 [m/s]$

the flux is

$\phi = 0$

now, the EMF is defined as:

$\epsilon =- \frac{d \phi}{dt}$ ,

if we consider $\Delta t= 2.4 [m/s]$ very small then we can re-write it as:

$\epsilon =- \frac{\Delta \phi}{\Delta t}$

$\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12$

then:

$\epsilon =- \frac{-0.12}{0.0024} = 50 [V]$

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

3 0

3 years ago

Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(

bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude