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3 years ago

HELP PLEASE!!!

Physics

1 answer:

Leya [2.2K]3 years ago

3 0

For the first one 320

second

1200W

Data

R = 12 Ω ∆V = 120V I =? P =?

Solution:

According to Ohm’s law,

∆V = I R

I = ∆V / R

= 120 / 12

= 10 A

Power P = I ∆V

= 10 x 120

= 1200 W

Third

∆V = 120 V P = 60 W I =? R =?

Use the formula, P = I ∆V

I = P / ∆V = 60 / 120 = 0.5 A

∆V = I R

R = ∆V / I = 120 / 0.5 = 240 Ω

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago

In a series RLC ac circuit, a second resistor is connected in series with the resistor previously in the circuit. As a result of

AnnyKZ [126]

Answer:

* The first thing we observe is that the frequency response does not change

* The current that circulates in the circuit decreases due to the new resistance at the resonance point,

Z = R + R₂

Explanation:

The impedance of a series circuit is

Z₀² = R² + (X_L-X_C) ²

when we place another resistor in series the initial resistance impedance changes to

Z² = (R + R₂) ² + (X_L - X_C) ²

let's analyze this expression

* The first thing we observe is that the frequency response does not change

* The current that circulates in the circuit decreases due to the new resistance at the resonance point,

Z = R + R₂

8 0

3 years ago

An unknown fluid flows at a speed of 31 m/s. Suppose the fluid has a mass of 47 kg runs at this speed. What is the fluid’s kinet

Leya [2.2K]

Answer:

22583.5J

Explanation:

KE=1/2 mv^2

=1/2*47Kg*(31m/s^2)

=23.5Kg * 961m/s^2

=22583.5J

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2 years ago

The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8

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Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) $\tau = \dfrac{\Delta L}{\Delta t}$

$\tau = \dfrac{0.8-3}{1.5}$

τ = -1.467 N m

b) angle at which fly wheel will turn

$\theta= \omega t +\dfrac{1}{2}\alpha t^2$

$\theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2$

$\theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2$

θ = 20.35 rad

c) work done on the wheel

W = τ x θ

W = -1.467 x 20.35 rad

W = -29.86 J

d) average power of wheel

$P_{av} =-\dfrac{W}{t}$

$P_{av} =-\dfrac{(-29.86)}{1.5}$

$P_{av} =19.91\ W$

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Answer:

Tension.

<em><u>tension</u></em> is the name of force that opposes or goes opposite of gravity

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