Electric potential, unlike electric potential energy, is measured in units of ________

Consider the Bose-Einstein distribution: F60 per unit energy can be written as N aE 6.) (20pts) Assuming BE <<1,obtain an

notsponge [240]

Answer:

Explanation:

Whacbdhnd

4 0

2 years ago

Before starting this problem, review Conceptual Example 3 in your text. Suppose that the hail described there comes straight dow

bulgar [2K]

Answer:

0.9 N

Explanation:

The force exerted on an object is related to its change in momentum by:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force exerted

$\Delta p$ is the change in momentum

$\Delta t$ is the time interval

The change in momentum can be rewritten as

$\Delta p = m(v-u)$

where

m is the mass

u is the initial velocity

v is the final velocity

So the formula can be rewritten as

$F=\frac{m(v-u)}{\Delta t}$

In this problem we have:

$\frac{m}{\Delta t}=0.030 kg/s$ is the mass rate

$u=-15 m/s$ is the initial velocity

$v=+15 m/s$ is the final velocity

Therefore, the force exerted by the hail on the roof is:

$F=(0.030)(+15-(-15))=0.9 N$

6 0

2 years ago

Calculate the resistance 50 feet of awg no. 12 solid copper wire round to nearest 0. 01 ohms.

emmasim [6.3K]

The resistance of the conductor is 0.07940 Ohms.

<h3>What is the relation between resistance and area of wire?</h3>

The wire's resistance is inversely related to its cross-sectional area; as the area drops, so does the resistance.

and it is formulated as:

$R=p\frac{l}{s}$

where,

<em>p </em>- resistivity of the conductor (0.0214-ohm mm²/m)

R - resistance

l- length of conductor (50 feet) (15.24 m)

s - the area of the wire

Thus the resistance can be calculated as

R = 0.07940 Ohms.

Learn more about resistance here:

brainly.com/question/11431009

#SPJ4

6 0

1 year ago

Which if, any, of these statements are true? (More than one may be true.) Assume the batteries are ideal. Check all that apply.

Maurinko [17]

Answer:

All are true except the last point that says that a battery is a current source and the current at the outlet is always the same.

Explanation:

A battery is an electro-chemical device which converts the chemical energy into usable electrical energy thus it provides electrical energy.
Since, the battery maintains a a constant potential difference between its terminals, once connected.
Since, the movement of electric current requires energy, which is supplied by the electric potential energy stored in the battery.
The current in the battery flows as per the Ohm's law and we can not say that the current leaving will always remain constant.
As the current is the flow of electric charge, and charges are not stored in batteries unlike capacitors, thus the current at the leaving end will depend on Ohm's law and will vary accordingly.

3 0

2 years ago

In a circus performance, a monkey on a sled is given an initial speed of 4.3 m/s up a 24◦ incline. The combined mass of the monk

Serggg [28]

Answer:

d = 1.15 m

Explanation:

In absence of friction, the change in kinetic energy of the combined mass of the monkey and the sled, must be equal (with opposite sign), to the change in gravitational potential energy:

ΔK = -ΔU

When friction is not negligible, the change in mechanical energy, must be equal to the work done by non-conservative forces (kinetic friction in this case):

ΔK + ΔU = Wnc (1)

As the monkey + sled reach to the maximum distance up the incline, they will come momentarily to a stop, so the final kinetic energy is 0.

$(K_{f} -K_{o}) = 0 - \frac{1}{2} * m*v_{o} ^{2} = -\frac{1}{2} *22.5kg*(4.3m/s)^{2} = -208.1J$

The change in gravitational energy, can be written as follows:

$(U_{f} - U_{o} ) = m*g*h - 0 = m*g*h = \\ \\ 22.5 kg*9.8 m/s2*d*sin (24 deg) = 89.7J*d$

The sum of these two quantities, must be equal to the work done by the friction force, along the distance d up the incline:

$W_{nc} = -\mu k*N*d$

The normal force, always normal to the surface, must be equal and opposite to the component of the weight normal to the incline:

$N = m*g*cos \theta = m*g*cos (24 deg) = \\ \\ 22.5 kg*9.8m/s2*0.913 = 201.4 N$

Replacing in the equation for Wnc:

$W_{nc} = -\mu k*N*d = -0.45*201.4 N*d = -90.6 N*d$

We can return to the equation (1) and solve for d:

$-208.1 J + 89.7N*d = -90.6N*d\\\\ d = \frac{208.1}{180.3} =1.15 m$

3 0

2 years ago

Electric potential, unlike electric potential energy, is measured in units of _________.