Is pressure defined as force per unit area

How do you find average velocity during free fall?

Genrish500 [490]

The sum of the initial and final velocity divided by 2.

6 0

3 years ago

Read 2 more answers

Does anyone know this?!

Valentin [98]

Answer:

2 is the numerical answer.

Explanation:

Hello there!

In this case, according to the given information and formula, it is possible for us to remember that equation for the calculation of the average kinetic energy of a gas is:

$KE=\frac{3}{2} \frac{R}{N_A} T$

Whereas R is the universal gas constant, NA the Avogadro's number and T the temperature.

Which means that for the given ratio, we can obtain the value as follows:

$=\frac{\frac{3}{2} \frac{R}{N_A} T_1}{\frac{3}{2} \frac{R}{N_A} T_2} \\\\=\frac{T_1}{T_2} \\\\=\frac{500K}{250K} \\\\=2$

Regards!

8 0

2 years ago

A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

$\mu$ = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N$

Magnitude of frictional force is 628.022466 N

$F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2$

$v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s$

Initial speed of the player is 8.61 m/s

4 0

2 years ago

Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%

djyliett [7]

Answer:

$6.8370869499\times 10^{20}\ N$

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

e = Charge of electron = $1.6\times 10^{-19}\ C$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

$q=imbalance\times n\times e$

Total number of protons and electrons in each sphere

$n=\dfrac{mN_AZe}{M}$

$q=imbalance\times \dfrac{mN_AZe}{M}$

$q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C$

$q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C$

Electrical force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N$

The electrostatic force of attraction between them is $6.8370869499\times 10^{20}\ N$

4 0

3 years ago

A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if

KengaRu [80]

Answer:

The value of the correct angle of banking for the road is $\theta = 67.76$ °

Explanation:

Given data

Velocity (v) = 60 $\frac{m}{s}$

Radius = 150 m

The velocity of the car in this case is given by

$v = \sqrt{r g \tan \theta}$

$v^{2} = r g \tan \theta$

$\tan \theta = \frac{v^{2} }{rg}$

Put all the values in above formula we get

$\tan \theta = \frac{60^{2} }{(150)(9.81)}$

$\tan \theta =$ 2.446

$\theta = 67.76$ °

Therefore the value of the correct angle of banking for the road is $\theta = 67.76$ °

4 0

2 years ago