Based on the trend in valence electrons across periods for main-group

Suppose you want to to double a copper wire's resistance. To what temperature, in degrees Celsius, must you raise it if it is or

sleet_krkn [62]

Answer:

T=280.41 °C

Explanation:

Given that

At T= 24°C Resistance =Ro

Lets take at temperature T resistance is 2Ro

We know that resistance R given as

R= Ro(1+αΔT)

R-Ro=Ro αΔT

For copper wire

α(coefficient of Resistance) = 3.9 x 10⁻³ /°C

Given that at temperature T

R= 2Ro

Now by putting the values

R-Ro=Ro αΔT

2Ro-Ro=Ro αΔT

1 = αΔT

1 = 3.9 x 10⁻³ x ΔT

ΔT = 256.41 °C

T- 24 = 256.41 °C

T=280.41 °C

So the final temperature is 280.41 °C.

8 0

3 years ago

State coulombs law in word

amid [387]

<h2>state coulombs law in word</h2>

: a statement in physics: the force of attraction or repulsion acting along a straight line between two electric charges is directly proportional to the product of the charges and inversely to the square of the distance between them

hope it helps

#carry on learning

3 0

3 years ago

A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert

pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

$\theta=65.4^{\circ}$

The projection of the length of the pendulum along the vertical direction is

$L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm$

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

$\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m$

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

$\Delta U = mg \Delta y$

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

$\Delta y = 0.211 m$ is the vertical displacement of the pendulum

So, the work done by gravity is

$W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J$

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

$\theta=90^{\circ}, cos \theta = 0$

and so the work done is zero.

5 0

3 years ago

A 0.50-kg bomb is sliding along an icy pond (frictionless surface) with a velocity of 2.0 m/s to the west. The bomb explodes int

nevsk [136]

Answer:

2.667m/s to the north and 3.333 m/s to the west

Explanation:

According to law of momentum conservation, the total momentum should be conserved before and after the explosion.

Before the explosion, the momentum was

0.5*2 = 1 kg m/s to the west

Therefore the total momentum after the explosion should be the same horizontally and vertically.

Vertically speaking, it was 0 before the explosion. After the explosion:

0.2*4 + 0.3v = 0

0.3v = -0.8

v = -0.8/0.3 = -2.667 m/s

So the vertical component of the 0.3kg piece is 2.667m/s to the north

Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:

0.2*0 + 0.3V = 1

0.3V = 1

V = 1/0.3 = 3.333 m/s

So the horizontal component of the 0.3kg piece is 3.333 m/s to the west

5 0

3 years ago

Read 2 more answers

What will happen to the astronaut when the jets produce these four forces: 10N, 10N, 9N, 9N?

sukhopar [10]

The astronaut would go the opposite direction due to Newton’s third law of -10N, -10N, -9N, -9N

Let me know if this helped you, please rank this was the brainlist answer if possible, thanks!

6 0

3 years ago