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3 years ago

How much energy does a pair of 800w hair straighteners transfer every second?

Physics

2 answers:

Solnce55 [7]3 years ago

5 0

Answer: 800 J

Explanation:

Power is the rate between the energy transferred (E) and the time taken (t):

$P=\frac{E}{t}$

In this problem, the hair straightener has a power of P = 800 W. therefore, the energy transferred in t = 1 second is

$E=Pt=(800 W)(1 s)=800 J$

babymother [125]3 years ago

3 0

Power is a measure of the rate of energy or work done on a system. It has units of energy per time. In SI, Joules per second and is equal to the units of Watts. So, for an 800 Watts hair straightener, the energy transfer per second would be 800 Joules.

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A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

$E_{1}+E_{3}-E_{2}=0$

We know that the electric field is:

$E=k\frac{q}{r^{2}}$

Where:

k is the Coulomb constant
q is the charge
r is the distance from the charge to the point

So, we have:

$k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0$

Let's solve it for r(3).

$\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0$

$r_{3}=0.0743\:$

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

3 0

2 years ago

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

marshall27 [118]

Answer:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Explanation:

Let the linear charge density of the charged wire is given as

$\frac{q}{L} = \lambda$

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

$\int E. dA = \frac{q}{\epsilon_0}$

here we have

$E \int dA = \frac{\lambda L}{\epsilon_0}$

$E. 2\pi r L = \frac{\lambda L}{\epsilon_0}$

so we have

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

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2 years ago

What part of your daily routine can be changed to help the climate change crisis?

Travka [436]

I recycle, would this help the climate?

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2 years ago

What is the energy released in this B- nuclear reaction 2K-> 2Ca0,e? (The atomic mass of 42 K is 41.962403 u and that of 42Ca

Katyanochek1 [597]

<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{19}^{42}\textrm{K}\rightarrow _{20}^{42}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{42}\textrm{K}$ = 41.962403 u

Mass of $_{20}^{42}\textrm{Ca}$ = 41.958618 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(41.962403-41.958618)=0.003785u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.003785u)\times c^2$

$E=(0.003785u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=3.526MeV$

Hence, the energy released in the given nuclear reaction is 3.526 MeV.

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3 years ago

Use the drop-down menus to identify the variables for the second hypothesis.

Debora [2.8K]

Answer: 1) independent

2) dependent

3) voltage

Explanation:

8 0

3 years ago

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