Does the temperature of the water change the fluid pressure in the water?

Select the decimal that is equivalent to 29/37

morpeh [17]

Answer:

0.78

Explanation:

6 0

2 years ago

An expensive vacuum system can achieve a pressure as low as 1.00×10^{-7} N/m^2 at 20ºC . How many atoms are there in a cubic cen

marta [7]

Answer:

$24.70818432141\times 10^7\ atoms$

Explanation:

P = Pressure = $1\times 10^{-7}\ N/m^2$

V = Volume = 1 cm³

n = Amount of substance

N = Number of atoms

$N_A$ = Avogadro's constant = $6.022\times 10^{23}\ /mol$

R = Gas constant = 8.314 J/k mol

T = Temperature = 273.15+20 = 293.15 K

From the ideal gas law

$PV=nRT\\\Rightarrow n=\frac{PV}{RT}$

$n=\frac{N}{N_A}$

$\frac{N}{N_A}=\frac{PV}{RT}\\\Rightarrow N=N_A\times \frac{PV}{RT}\\\Rightarrow N=\frac{1\times 10^{-7}\times 1\times 10^{-6}}{8.314\times 293.15}\times 6.022\times 10^{23}\\\Rightarrow N=24708184.32141\ atoms=24.70818432141\times 10^7\ atoms$

The number of atoms is $24.70818432141\times 10^7\ atoms$

8 0

3 years ago

Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object

Nookie1986 [14]

Explanation:

We have,

The initial position of an object is zero.

The starting velocity is 3 m/s and the final velocity was 10 m/s.

The object moves with constant acceleration..

The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".

8 0

3 years ago

A soccer ball collides with another soccer ball at rest. the total momentum of the balls

Juliette [100K]

I am pretty sure the answer is C.

4 0

3 years ago

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$