When a force is applied through a distance , work is done

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

2 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

Pls someone I need it urgently and explain Solving and explanation so I can understand Thank you

Temka [501]

Answer:

f = 6.37 Hz, T = 0.157 s

Explanation:

The expression you have is

y = 5 sin (3x - 40t)

this is the equation of a traveling wave, the general form of the expression is

y = A sin (kx - wt)

where A is the amplitude of the motion, k the wave vector and w the angular velocity

Angle velocity and frequency are related

w = 2π f

f = w / 2π

from the equation w = 40 rad / s

f = 40 / 2π

f = 6.37 Hz

frequency and period are related

f = 1 / T

T = 1 / f

T = 1 / 6.37

T = 0.157 s

4 0

3 years ago

Which of the following would have more momentum than fiona, a 313kg hippo, running at 8m/s?

kap26 [50]

Let's calculate the momentum of Fiona, given by the product between its mass and its speed:
$p=mv=(313 kg)(8 m/s)=2504 kg m/s$

Now let's compare it with the momentum of the other animals:
a) the mass of the sea turtle is missing, so we cannot calculate its momentum.
b) the momentum of the dolphin is
$p=mv=(150 kg)(18 m/s)=2700 kg m/s$
c) the momentum of the horse is
$p=mv=(380 kg)(4 m/s)=1520 kg m/s$
d) the momentum of the lion is
$p=mv=(190 kg)(11 m/s)=2090 m/s$

And we can see that the correct answer is b), because the momentum of the dolphin is greater than the momentum of Fiona.

5 0

3 years ago

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be

Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

F = $m \times a$