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3 years ago

Calculate the resistance of a 900.-watt toaster

2 answers:

8 0

Using the Definition of Potency, we have:

$P= \frac{V^2}{R}$

Entering the unknowns:

$P= \frac{V^2}{R} \\ 900*W= \frac{(120*V)^2}{R} \\ R= \frac{14400*V^2}{900*V*A} \\ R=16* \frac{V}{A} \\ \boxed {R=16Ohm}$

If you notice any mistake in my english, please let me know, because i am not native.

liubo4ka [24]3 years ago

5 0

Answer : Resistance is 16 ohms.

Explanation :

It is given that,

Power of toaster, $P=900\ Watt$

Voltage, $V=120\ V$

According to ohm's law, V = I R

Mathematically power is defined as :

$P=\dfrac{V^2}{R}$

$R=\dfrac{V^2}{P}$

$R=\dfrac{(120\ V)^2}{900\ W}$

$R=16\ \Omega$

So, the resistance of the toaster is 16 ohms.

Hence, this is the required solution.

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The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of th

SVETLANKA909090 [29]

Answer:

$F_b$ = 1647.92 N

Explanation:

Given:

length of the arm, r = 0.465 m

distance of forearm from elbow, r' = 2.15 cm = 0.0215 m

Mass of the forearm, M = 2.45 kg

Mass of the object, m = 6.55 kg

Let, the Force by bicep be, $F_b$

under the motionless condition, the net moment about elbow is zero

thus,

$F_b$ × 0.0215 - Mg × (0.465/2) - mg × (0.465) = 0

$F_b$ × 0.0215 - 2.45 × 9.8 × (0.465/2) - 6.55 × 9.8 × (0.465) = 0

$F_b$ × 0.0215 - 5.582 - 29.848 = 0

$F_b$ = 1647.92 N

hence, the force exerted by the elbow is 1647.92 N

4 0

4 years ago

Read 2 more answers

Colonel John P. Stapp, USAF, participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, h

borishaifa [10]

we assume the acceleration is constant. we choose the initial and final points 1.40s apart, bracketing the slowing-down process. then we have a straightforward problem about a particle under constant acceleration. the initial velocity is v xi =632mi/h=632mi/h( 1mi 1609m )( 3600s 1h )=282m/s (a) taking v xf =v xi +a x t with v xf =0 a x = t v xf −v xf = 1.40s 0−282m/s =−202m/s 2 this has a magnitude of approximately 20g (b) similarly x f −x i = 2 1 (v xi +v xf )t= 2 1 (282m/s+0)(1.40s)=198m

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3 years ago

A. Draw the electric field lines around a negative charge.

Alborosie

<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric fields, so this helps us to see the problem more real. So an electric field line is an imaginary line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. </em>This can be expressed as follows:

$F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m$

Then:

$F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}$

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

$q_{1} \ and \ q_{2}$

If the amount of charge on one of the objects is tripled, let's say this is the charge $q_{2}$ , then the new charge is:

$q_{N}=3q_{2}$

In the formula of Coulomb:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}$

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2}$

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod repels positive charges on the sphere, creating zones of negative and positive charge as indicated in the second Figure.

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3 years ago

Read 2 more answers

What is the electric potential 12 cm away from a charge of –5.2 × 10^–6C?

MAXImum [283]

The equation for electric potential of a point charge is:

V=(k*q)/r where k=9*10^11 N m²/C², q is the charge and r is the distance from the charge to the point in which we are calculating the potential.

q=-5.2*10^-6 C
r=12 cm = 0.12 m

Now we plug in the numbers and get:

V=-3.9*10^5 Volts, so the correct answer is the second one.

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3 years ago

Two identical guitar strings are prepared such that they have the same length ( 0.62 m ) and are under the same amount of tensio

Tanya [424]

Answer:

The speed is 401.76m/s

Explanation:

To find the fundamental frequency we use

3rd.harmonic - 2nd.harmonic = 324Hz

And we know that

f=v/2L

So v=f2L

v= 324 x 2 x 0.62= 401.76m/s

4 0

3 years ago