All categories

2 years ago

Please help help help mee

Physics

1 answer:

vampirchik [111]2 years ago

4 0

Answer:

.60 A

Explanation:

The RMS (Root Mean Square) voltage (the effective voltage) of the PEAK voltage is .7 * peak

.7 * 425 = 297.5 v

Now, V = IR re-arranged to V/R = I

297.5 v / 500 ohm = .595 A = ~ .6 A

You might be interested in

A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at

storchak [24]

Answer

given,

wavelength = λ = 18.7 cm

= 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A) angular frequency = ?

$f = \dfrac{v}{\lambda}$

$f = \dfrac{0.38}{0.187}$

$f =2.03\ Hz$

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

$K = \dfrac{2\pi}{\lambda}$

$K= \dfrac{2\pi}{0.187}$

$K =33.59\ m^{-1}$

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k x - ω t)

y ( x ,t) = 2.34 x sin(33.59 x - 12.75 t)

4 0

3 years ago

A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir

Pachacha [2.7K]

Answer:

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

$\vec{F} = q \vec{E}$

$11\hat{i} = -5 \times 10^{-6} \vec{E}$

$\vec{E} =-2.2 \times 10^{6}\hat{i}$

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

$\vec{F} = q \vec{E}$

$\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}$

$\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

Force experienced by the photon in the same field is equal to $\vec{F}= -3.52\times 10^{-13}\hat{i}\ N$

3 0

3 years ago

Someone help me please

ivanzaharov [21]

C. cooked noodles and water
because noodles are long and water has no shape or size.
if you have any problems with this answer,
comment and I will fix it.
Thank s!

6 0

3 years ago

Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimu

Naddika [18.5K]

In the magnitud is in good

7 0

3 years ago

Read 2 more answers

The acceleration due to gravity on the Moon's surface is

Molodets [167]

Answer:

50 lb

Explanation:

Given,

The weight of astronaut's life support backpack on Earth (w) = 300 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

$g'=\frac{g}{6}$

We know that weight of an object on Earth is,

$w = m\times g$

$m = \frac{w}{g}$

Similarly, weight on Moon will be

$w' = m\times g'$

$w' = \frac{w}{g}\times\frac{g}{6}$

$w' = \frac{300}{6}$

$w' = 50$

Thus the astronaut's life support backpack will weigh 50 lb on Moon.

7 0

3 years ago