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4 years ago

An electric heater uses 3/ 5

2 answers:

5 0

In this question, all the vital information's required to reach the solution is already given. Based on those information's the answer to the question can be easily reached. Let us first write down all the information's given.
Amount of power used by the electric heater in 1 hour = 3/5 watts
Then
Amount of power used by the electric heater in 1/2 hour = (3/5) * (1/2) watts
= (3 * 1)/(5 * 2) watts
= 3/10 watts
So 3/10 watts of power is required by the electric heater in 1/2 an hour. The correct option among all the options given in the question is option "a".

Andru [333]4 years ago

3 0

This question could not come from any school source. If it did,
then you should always be suspicious of anything that comes
from that source. The question contains a fundamental mistake,
and misleads the student.

"Power", measured in "watts", is a rate, not a quantity. It tells
how fast you're using energy. You don't use some watts this
morning and some more watts this afternoon. That's like saying
that you'll drive some miles per hour today, and then you'll drive
some more miles per hour tomorrow. It doesn't work that way.

The heater doesn't use up some amount of watts. It uses up
some amount of energy. "Watts" is the measure of how fast
or slowly it uses up the energy.

This question makes sense if you get rid of the 'watts' and call it
'pints of milk' instead.

   "My cats drink 3/5 pint of milk each hour.
      How many pints of milk will they drink in 1/2 hour ?"

   (1/2 hour) x (3/5 pint/hour)

   = ( 1/2 x 3/5) (pint)

   =    3/10 pint .

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4 years ago

A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R. The centra

Dmitriy789 [7]

Answer:

(a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

We need to calculate the electric field due to the rod

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

$E=0$

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}$

$E\propto\dfrac{1}{z^2}$

(c). In terms of R,

We need to calculate the positive distance

If $E\rightarrow E_{max}$

Then, $\dfrac{dE}{dz}=0$

$\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0$

Taking only positive distance

$z=\dfrac{R}{\sqrt{2}}$

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

$E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

$E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}$

$E_{max}=15418.7\ N/C$

$E_{max}=1.54\times10^{4}\ N/C$

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

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3 years ago

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3 years ago

Read 2 more answers

A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w

inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is $v=\frac{d}{t}$ , where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is $d=v\cdot t$

Now, the distance equation for the man would be:

$d_{man}=v_{man}\cdot t=1.6\cdot t$

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

$d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)$