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AysviL [449]
3 years ago
7

You and a friend are studying late at night. There are three 110 W light bulbs and a radio with an internal resistance of 56.0 Ω

plugged into the living room electrical outlets. You decide to have a break and have a snack, turning on the pop-corn machine that draws 7.00 A. How much electrical power are you using? Data: Assume the electric company provide you with 110 V potential difference.
Physics
1 answer:
Sliva [168]3 years ago
8 0

Answer:

The total electrical power we are using is: 1316 W.

Explanation:

Using the ohm´s law V=I*R and the formula for calculate the electrical power, we can find the total electrical power that we are using. First we need to find each electrical power that is using every single component, so the radio power is:I=\frac{V}{R}=\frac{110 (v)}{56(ohms)}=1.96(A), so the radio power is: P=I*V=1.96(A)*110(v)=216(W), then we find the pop-corn machine power as: P=I*V=7(A)*110(v)=770(W) and finally there are three light bulbs of 110(W) so: P=3*110(W)=330(W) and the total electrical power is the adding up every single power so that: P=330(W)+770(W)+216(W)=1316(W).

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Which of the following statements are true for magnetic force acting on a current-carrying wire in a uniform magnetic field? Che
qaws [65]

Answer:

The following statements are correct.

1. The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

2. The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

3. The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

Wrong statements:

1. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic field lines.

Explanation:

6 0
3 years ago
A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a
iren [92.7K]

Answer:

√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

3x=(1/2)a(t')^2

Solving for t':

t'=√(6x/a)

Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

Evaluating at t=t'

v(t')=a*√(6x/a)=√(6*x*a)

Which is the final velocity

Have a nice day!

3 0
3 years ago
Does a set of all sets contain itself?
ipn [44]
The set of all sets that are not members of themselves. This contradiction is Russell's paradox.
5 0
3 years ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
pickupchik [31]
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height. 

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
2 years ago
when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why​
Lady_Fox [76]

Answer:

hsvshxansjusjsnwjwisks

4 0
3 years ago
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