By standard convention, both the electric potential and the the electric potential energy between two charges is taken to be zer

In a given city, the permissible limit of CO (carbon monoxide) in the air is 100 parts per million (ppm). The city monitors the

Sav [3153]

The city evaluates the continuous increase of carbon monoxide from different origins each year. According to calculations, in the year "C: 2019"<span> (rounded to the closest whole number), the concentration of CO will surpass the allowed threshold.

If this is not correct, feel free to inform me and I will find out the right answer. However, I am confident this is accurate.:) </span>

8 0

2 months ago

Read 2 more answers

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Sav [3153]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

$a=\frac{v^2}{R}$

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = $9.81\ m/s^2$

v = ?

R = 13.2 m

Por lo tanto,

$9.81=\frac{v^2}{13.2}$

$v^2=9.81\times {13.2}$

v = 11.4 m/s

8 0

3 months ago

If the wire is replaced by an infinite current sheet with density Js = 0.40 A/m, what would be the magnetic flux (in T · m2) thr

Sav [3153]

Answer:

Clarification: $\phi _{B} =0.855 T-m^{-2}$

Considering the following data the density of the current sheet is 0.40 A/m

The length a = 0.27 m

and the width b = 0.63 m

For an infinite sheet, the magnetic field is described as

The magnetic flux is given as

$B = \mu _{O}J$

magnetic flux =

$\phi _{B} = BA$

$= \mu _{O}Jab$

$= 4\pi *0.40*0.27*0.63$

$\phi _{B} =0.855 T-m^{-2}$

6 0

1 month ago

The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

inna [3103]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The velocity of the center of mass for the two-block system just prior to their collision is 2.9489 m/s

Explanation:

Provided information:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = collision time = 0.5 s

Question: What is the velocity of the center of mass for the two-block system right before the blocks collide, vf =?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It’s essential to calculate the required force:

$F=(m+m)*g*sin\theta$

Here, g = acceleration due to gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

6 0

2 months ago

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

Keith_Richards [3271]

Answer:

The flux across the cube's surface is $2.314\ Nm^{2}/C$ .

Solution:

According to the details provided:

Cube edge length, a = 8.0 cm = $8.0\times 10^{- 2}\ m$ .

Volume charge density, $\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}$ .

Now,

To find the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$

where

$\phi$ = electric flux

$\epsilon_{o} = 8.85\times 10^{- 12}\ F/m$ = permittivity of vacuum.

The volume charge density for this scenario is described by:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$

Cube volume, $V = a^{3}$ .

Thus,

$V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}$ .

The total charge can be derived from equation (2):

$q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}$ .

$q = 2.048\times 10^{-11}\ F = 20.48\ pF$ .

Now, insert the value of 'q' into equation (1):

$\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C$ .

5 0

2 months ago