The speed of a motorcycle at a particular moment is it’s average speed or is instantaneous speed?

Carbon is allowed to diffuse through a steel plate 11 mm thick. The concentrations of carbon at the two faces are 0.88 and 0.41

Serhud [2]

Answer:

The temperature is 2584.5 K

Explanation:

Given:

Activation energy $Q = 80000$ $\frac{J}{mol}$

Preexponential $D= 6.2 \times 10^{-7}$ $\frac{m^{2} }{s}$

Diffusion flux $J = 6.4 \times 10^{-10}$ $\frac{kg}{m^{2} s}$

Thickness of plate $\Delta x = 11 \times 10^{-3}$ m

Concentration of carbon at two faces $\Delta C = (0.88 - 0.41 ) = 0.47$ $\frac{kg}{m^{3} }$

From the formula of temperature in terms of diffusion flux,

$T = (\frac{Q}{R} ) \frac{1}{\ln (\frac{D\Delta C}{J\Delta x} )}$

Where $R =$ 8.314 $\frac{J}{mol.K}$ ( gas constant )

Put the values and find the temperature,

$T = (\frac{80000}{8.314} ) \frac{1}{\ln (\frac{6.2 \times 10^{-7} \times 0.47 }{6.4 \times 10^{-10}\times 11 \times 10^{-3} } )}$

$T = 2584.5$ K

Therefore, the temperature is 2584.5 K

8 0

3 years ago

The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

love history [14]

Answer:

Part a)

$EMF = 14 \times 10^{-3} V$

Part b)

$EMF = 15.67 \times 10^{-3} V$

Explanation:

As we know that magnetic flux through the loop is given as

$\phi = B.A$

now we have

$\phi = B\pi r^2$

now rate of change in flux is given as

$\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}$

now we know that

$A = \pi r^2$

$0.285 = \pi r^2$

$r = 0.30 m$

Now plug in all data

$EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)$

$EMF = 14 \times 10^{-3} V$

Part b)

Now the radius of the loop after t = 1 s

$r_1 = r_0 + \frac{dr}{dt}$

$r_1 = 0.30 + 0.037$

$r_1 = 0.337 m$

Now plug in data in above equation

$EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)$

$EMF = 15.67 \times 10^{-3} V$

5 0

3 years ago

Blocks A and B are identical metal blocks. Initially block A is neutral, and block B has a net charge of 7.0 nC. Using insulatin

maxonik [38]

Answer:

Block A will have a final charge of 3.5nC.

Explanation:

This is because at the point of contact with Block B, which is electrically positive, the electrons in Block A will be attracted to the excess 'unpaired' protons in block B. Hence, the electrons will flow into Block B causing unpaired protons to remain in Block A.

This process is called Charging by Conduction.

This charging process will continue until the charges are evenly distributed between both objects.

In case you're wondering, "<em>how's all this possible within a few seconds</em>?", remember that electrons travel very fast and so, this process is a rather rapid one.

6 0

3 years ago

Read 2 more answers

Scenario

Anvisha [2.4K]

Answer:

1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

0 = y₀ + 0 - ½ g t2

t = $\sqrt{ \frac{2y_o}{g} }$

t = √(2 2650/9.8)

t = 23.26 s

this is the horizontal scrolling time

x = v₀ t

x = 366.67 23.26

x = 8527 m

the speed at the point of arrival is

v_y = v_{oy} - g t = 0 - gt

v_y = - 9.8 23.26

v_y = -227.95 m / s

Module and angle form

v = $\sqrt{v_x^2 + v_y^2}$

v = √(366.67² + 227.95²)

v = 431.75 m / s

θ = tan⁻¹ (v_y / vₓ)

θ = tan⁻¹ (227.95 / 366.67)

θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

cos θ = v₀ₓ / v₀

sin θ = v_{oy} / v₀

v₀ₓ = v₀ cos θ

v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

y = y₀ + v_{oy} t - ½ g t²

0 = y₀ + vo sin θ t - ½ g t²

0 = -50 + vo sin 50 t - ½ 9.8 t²

x = x₀ + v₀ₓ t

0 = x₀ + vo cos θ t

0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

50 = 0,766 vo t – 4,9 t²

400 = 0,643 vo t

resolved

50 = 0,766 ( $\frac{400}{0.643 \ t}$ ) t – 4,9 t²

50 = 476,52 t – 4,9 t²

t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

t = [97.25 ± $\sqrt{97.25^2 - 4 \ 10.2}$ ] / 2

t = 97.25 ±97.04] 2

t₁ = 97.145 s

t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

t = 97.145 s

let's find the initial velocity

x = x₀ + v₀ cos 50 t

0 = -400 + v₀ cos 50 97.145

v₀ = 400 / 62.44

v₀ = 6.4 m / s

5 0

2 years ago

Would the springs inside a bathroom scale be more compressed or less compressed if you weighed yourself in an elevator that was

Ber [7]

More compressed. moving up = apparent weight (i.e., your norma force) is greater. this means you’ll weighr more and push those springs down even more than you would at rest.

6 0

2 years ago