Which layer is made of flowing solid rock

A car starting from rest (i.e. initial velocity = 0.0 m/s), moves in the positive X-direction with a constant average accelerati

olga_2 [115]

Answer:

Final speed of the car, v = 24.49 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 0

Acceleration, $a=3.1\ m/s^2$

Time taken, t = 7.9 s

We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :

$v=u+at$

$v=0+3.1\times 7.9$

v = 24.49 m/s

So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.

8 0

3 years ago

A circuit is built based on the circuit diagram shown.

julsineya [31]

The current in the 50 Ω resistor is A) 1.2 A

4 0

3 years ago

Read 2 more answers

If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is

leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

$\rm y = 0.02\sin(30x-200 t).$

The vertical displacement of a wave is given in generalized form as

$\rm y = A\sin(kx -\omega t).$

<em>where</em>,

A = amplitude of the displacement of the wave.
k = wave number of the wave = $\dfrac{2\pi }{\lambda}.$
$\lambda$ = wavelength of the wave.
x = horizontal displacement of the wave.
$\omega$ = angular frequency of the wave = $\rm 2\pi f$ .
f = frequency of the wave.
t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

$\rm A=0.02.\\k=30.\\\omega =200.\\$

therefore,

$\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.$

It is the required frequency of the wave.

3 0

3 years ago

Consider a 20 cm thick granite wall with a thermal conductivity of 2.79 W/m·K. The temperature of the left surface is held const

kozerog [31]

Answer:

The right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

Explanation:

Thickness of the wall is L= 20cm = 0.2m

Thermal conductivity of the wall is K = 2.79 W/m·K

Temperature at the left side surface is T₁ = 50°C

Temperature of the air is T = 22°C

Convection heat transfer coefficient is h = 15 W/m2·K

Heat conduction process through wall is equal to the heat convection process so

$Q_{conduction} = Q_{convection}$

Expression for the heat conduction process is

$Q_{conduction} = \frac{K(T_1 - T)}{L}$

Expression for the heat convection process is

$Q_{convection} = h(T_2 - T)$

Substitute the expressions of conduction and convection in equation above

$Q_{conduction} = Q_{convection}$

$\frac{K(T_1 - T_2)}{L} = h(T_2 - T)$

Substitute the values in above equation

$\frac{2.79(50- T_2)}{0.2} = 15(T_2 - 22)\\\\T_2 = 35.5^\circC$

Now heat flux through the wall can be calculated as

$q_{flux} = Q_{conduction} \\\\q_{flux} = \frac{K(T_1 - T_2)}{L}\\\\q_{flux} = \frac{2.79(50 - 35.5)}{0.2}\\\\q_{flux} = 202.3W/m^2$

Thus, the right wall surface temperature and heat flux through the wall is 35.5°C and 202.3W/m²

6 0

3 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago