How do you say scuba dive in spanish

The wires in a household lamp cord are typically 3.5 mm apart center to center and carry equal currents in opposite directions.

zhuklara [117]

Explanation:

It is given that,

Distance between wires, d = 3.5 mm = 0.0035 m

Power of light bulb, P = 100 W

Potential difference, V = 120 V

(a) We need to find the force per unit length each wire of the cord exert on the other. It is given by :

$\dfrac{F}{l}=\dfrac{\mu_o I^2}{2\pi r}$

Power, P = V × I

$I=\dfrac{P}{V}=\dfrac{100}{120}=0.83\ A$

This gives, $\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times (0.83)^2}{2\pi \times 0.0035}$

$\dfrac{F}{l}=0.0000393\ N/m$

$\dfrac{F}{l}=3.93\times 10^{-5}\ N/m$

(b) Since, the two wires carry equal currents in opposite directions. So, teh force is repulsive.

(c) This force is negligible.

Hence, this is the required solution.

8 0

2 years ago

Plants make ___________ using carbon dioxide, water and energy from the sun.

gizmo_the_mogwai [7]

Answer:

I think it's carbohydrates

Explanation:

The energy from light causes a chemical reaction that breaks down the molecules of carbon dioxide and water and reorganizes them to make the sugar (glucose) and oxygen gas.

7 0

2 years ago

You wish to cook some pasta and so take some water and heat it on a stove. If you use 2000 grams of water with an initial temper

Ronch [10]

Answer:

the stove energy went into heating water is 837.2 kJ.

Explanation:

given,

mass of water = 2000 grams

initial temperature = 0° C

Final temperature = 100° C

specific heat of water (c) = 4.186 joule/gram

energy = m c Δ T

= 2000 × 4.186 × (100° - 0°)

= 837200 J

= 837.2 kJ

hence, the stove energy went into heating water is 837.2 kJ.

3 0

2 years ago

If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to

-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

$\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm$

Moment of inertia from neutral axis will be given by $\frac {bd^{3}}{12}+ ay^{2}$

Therefore, moment of inertia will be

$\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}$

Bending stress at top= $\frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa$

Bending stress at bottom= $\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03$ Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

8 0

2 years ago

What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?

MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

$I = \frac{P}{A}$