If 0.8 Wh of electrical energy is lost as heat, how much heat energy (in Btu) is produced?

Lilli suggests that they explore the simulation starting with varying only a single parameter in order to understand the role of

mrs_skeptik [129]

Answer:

B.

Explanation:

One of the ways to address this issue is through the options given by the statement. The concepts related to the continuity equation and the Bernoulli equation.

Through these two equations it is possible to observe the behavior of the fluid, specifically the velocity at a constant height.

By definition the equation of continuity is,

$A_1V_1=A_2V_2$

In the problem $A_2$ is $2A_1$ , then

$A_1V_1=2A_1V_2$

$V_2 = \frac{V_1}{2}$

Here we can conclude that by means of the continuity when increasing the Area, a decrease will be obtained - in the diminished times in the area - in the speed.

For the particular case of Bernoulli we have to

$P_1 + \frac{1}{2}\rho V_1^2 = P_2 +\frac{1}{2}\rho V_2^2$

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-V_2^2)$

For the previous definition we can now replace,

$P_2-P_1 = \frac{1}{2} \rho (V_1^2-(\frac{V_1}{2})^2)$

$\Delta P = \frac{3}{8} \rho V_1^2$

Expressed from Bernoulli's equation we can identify that the greater the change that exists in pressure, fluid velocity will tend to decrease

The correct answer is B: "If we increase A2 then by the continuity equation the speed of the fluid should decrease. Bernoulli's equation then shows that if the velocity of the fluid decreases (at constant height conditions) then the pressure of the fluid should increase"

4 0

2 years ago

If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it

Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

$\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

Explanation:

From the question we are told that

The mass of the rocket is m = 6300 kg

The velocity at gas is being ejected is u = 2000 m/s

The initial acceleration desired is $a = 27.0 \ m/s$

The time taken for the gas to be ejected is t = 1 s

Generally this desired acceleration is mathematically represented as

$a = \frac{u * \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}$

Here $\frac{\Delta m}{\Delta t }$ is the rate at which gas is being ejected with respect to time

Substituting values

$27 = \frac{2000 * \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}$

=> $170100 -27* \frac{\Delta m}{\Delta t} = 2000 * \frac{\Delta m}{\Delta t}$

=> $170100 = 2027 * \frac{\Delta m}{\Delta t}$

=> $\frac{\Delta m}{\Delta t} = \frac{170100}{2027}$

=> $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

3 0

2 years ago

Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to anothe

levacccp [35]

Answer:

Obviously Lengthen... $T = 2\pi \sqrt{L/g}$ or $g = 4\pi ^{2} L/g$

Explanation:

As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...

4 0

2 years ago

You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current i

Anni [7]

Answer:

radius of the loop = 7.9 mm

number of turns N ≅ 399 turns

Explanation:

length of wire L= 2 m

field strength B = 3 mT = 0.003 T

current I = 12 A

recall that field strength B = μnI

where n is the turn per unit length

vacuum permeability μ = $4\pi *10^{-7} T-m/A$ = 1.256 x 10^-6 T-m/A

imputing values, we have

0.003 = 1.256 x 10^−6 x n x 12

0.003 = 1.507 x 10^-5 x n

n = 199.07 turns per unit length

for a length of 2 m,

number of loop N = 2 x 199.07 = 398.14 ≅ 399 turns

since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.

this length is also equal to the circumference of each loop

the circumference of each loop = $2\pi r$

0.005 = 2 x 3.142 x r

r = 0.005/6.284 = $7.9*10^{-4} m$ = 0.0079 m = 7.9 mm

8 0

2 years ago

Tarzan and Jane. Because of your concern that incorrect science is being taught to children when they watch cartoons on TV, you

creativ13 [48]

Answer:

The maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

Explanation:

Let

m = Mass of Tarzan

M = Mass of Jane

Given

M = 2m

To calculate the maximum height Tarzan and Jane can swing, we make use of the potential energy at their initial and final position.

Reason being that;

At both the initial and final position, velocity is 0, so there's no kinetic energy.

And the potential energy remains the same (i.e constant) at any given point in the system.

Using P.E = mgh.

At initial position, PE1 = mgh

At final position, PE2 = (m + M)gH.

Where h and H represent the initial and final heights.

m + M is the new weight after Jane and Tarzan swing

Equating PE1 to PE2

mgh = (m + M)gH

By substituton (M = 2m)

mgh = (m + 2m)gH

mgh = 3mgH

Make H the subject of the formula

H = mgh/3mg

H = ⅓h

Hence, the maximum height Tarzan and Jane can swing as a fraction of her initial heigh is ⅓h

From the question, the new height looks to be about ½ that of Jane's original position; i.e. ½h

The calculated height is smaller than what the cartoon is showing;

We can conclude that the cartoon is wrong.

4 0

2 years ago