The first thing you should know is that the friction force is equal to the coefficient of friction due to normal force.
Therefore, clearing the normal force we have:
The friction is 565N.
(565 / 0.8) = 706.25N. weight.

6 0

3 years ago

How much energy (in joules) is required to evaporate 0.0005 kg of liquid ammonia to vapor at the same temperature?

Lana71 [14]

Answer:

685.6 J

Explanation:

The latent heat of vaporization of ammonia is

L = 1371.2 kJ/kg

mass of ammonia, m = 0.0005 Kg

Heat = mass x latent heat of vaporization

H = 0.0005 x 1371.2

H = 0.686 kJ

H = 685.6 J

Thus, the amount of heat required to vaporize the ammonia is 685.6 J.

6 0

2 years ago

In some creatures, bioluminescence transforms

8090 [49]

OB which is light and heat

3 0

3 years ago

Read 2 more answers

An airplane wing is designed so that the speed of the air across the top of the wing is 297 m/s when the speed of the air below

Crank

Answer:

The value is $F_L = 759200 \ N$

Explanation:

From the question we are told that

The speed of air across the top of its wings is $v = 297 \ m/s$

The speed of air below its wings is $u = 209 \ m/s$

The density of air is $\rho = 1.29 \ kg/m^3$

The area of the wing is $A = 26.0 \ m^2$

Generally the lifting force is mathematically represented as

$F_L = \Delta P * A$

Here $\Delta P$ is the difference in kinetic energy density between the top and the bottom

$\Delta P = \frac{1}{2} * \rho * [v^2 - u^2 ]$

=> $\Delta P = \frac{1}{2} * 1.29 * [297^2 - 209^2 ]$

=> $\Delta P = 2.92 *10^{4} \ Pa$

$F_L = 2.92 *10^{4} * 26.0$

=> $F_L = 759200 \ N$

3 0

3 years ago