Can someone please help me with this problem. I’ve tried but I can’t seem to figure it out.

Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

Travka [436]

Answer:

22.1 V

Explanation:

We are given that

$A_1=0.7 cm^2=0.7\times 10^{-4} m^2$

$A_2=2.5 cm^2=2.5\times 10^{-4} m^2$

$A_3=2.2 cm^2=2.2\times 10^{-4} m^2$

$A_4=3 cm^2=3\times 10^{-4} m^2$

Using $1cm^2=10^{-4} m^2$

We know that

$R=\frac{\rho l}{A}$

In series

$R=R_1+R_2+R_3+R_4$

$R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}$

$R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}$

Substitute the values

$R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})$

$R=\rho l(2.62\times 10^4)$

$V=145 V$

$I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}$

Voltage across the 2.5 square cm wire= $IR=I\times \frac{\rho l}{A_2}$

Voltage across the 2.5 square cm wire= $\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V$

Voltage across the 2.5 square cm wire=22.1 V

6 0

3 years ago

A solid block, with a mass of 0.15kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring const

iren [92.7K]

Answer:

16.1 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

$U=\frac{1}{2}kx^2$

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

$K=\frac{1}{2}mv^2$

where m = 0.15 kg is the mass of the block and v is its speed.

Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:

$\frac{1}{2}kx^2=\frac{1}{2}mv^2\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(316 N/m)(0.35 m)^2}{0.15 kg}}=16.1 m/s$

4 0

3 years ago

A 0.38 kg drinking glass is filled with a hot liquid. The liquid transfers 7032 J of energy to the glass. If the

kaheart [24]

Answer:

841 J/kg.K

Explanation:

The computation of the specific hear of the glass is shown below:

As we know that

E= cmΔt

where

c denotes specific heat

m denotes 0.38 kg

Δt = temperature = 22k

E denotes energy = 7032 J

Now

7032 J = (0.38) (22) (c)

7032 J = 8.36 (c)

So C = 7032 J ÷ 8.36

= 841 J/kg.K

7 0

2 years ago

If the archerfish spits its water 60. degrees from the horizontal aiming at an insect 2.0 m above the surface of the water, how

MA_775_DIABLO [31]

<span>Using the kinematic equations below, we can calculate the initial velocity required. Angle of projectile = 60 degrees Acceleration due to gravity (Ay) = -10 m/s^2 (negative because downward) Height of projectile (Dy) = 2m Vfy^2=Voy^2 +2*Ay*Dy Vfy = 0 m/s because the vertical velocity slows to zero at the height of its trajection. So... 0 = Voy^2 + 2(-10)(2) 0 = Voy^2 - 40 40 = Voy^2 Sqrt40 = Voy 6.32 m/s = Voy THIS IS NOT THE ANSWER. THIS IS JUST THE INITIAL VELOCITY IN THE Y DIRECTION. Using trigonometry, Tan 60 = Voy/Vox. Tan 60 = 6.32/Vox. Vox*Tan 60 = Vox Vox = 10.95 m/s. Now, using Vox = 10.95 and Voy = 6.32, we can use pythagorean theorem to find the total Vo. A^2 +B^2 = C^2 10.95^2 + 6.32^2 = C^2 Solving for C = 12.64 m/s This is the velocity required to hit the surface. You can also calculate a bunch of other stuff now using the other kinematic equations. V = 12.64 m/s</span>

5 0

3 years ago

Read 2 more answers

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

Damm [24]

Answer:

$0.05\ \text{kJ/kg}$