Just wanna make sure im right

What is a substance?

jeka94

Answer:

a mixture that can be separated

6 0

2 years ago

Please help 2.

Mariulka [41]

Answer:

The new volume of the balloon is 8.8 L

Explanation:

Charles's Law consists of the relationship between the volume and the temperature of a certain amount of ideal gas, which is kept at a constant pressure. This law establishes that the volume is directly proportional to the temperature of the gas, that is to say that if the temperature increases, the volume of the gas increases and that on the contrary if the temperature of the gas decreases, the volume decreases.

In other words, Charles's law states that when the amount of gas and pressure are kept constant, the ratio between the volume and the temperature will always have the same value:

$\frac{V}{T}=k$

Having a certain volume of gas V1 that is at a temperature T1, by varying the volume of gas to a new value V2, then the temperature will change to T2, and it will be fulfilled:

$\frac{V1}{T1}=\frac{V2}{T2}$

In this case:

V1= 8.5 L
T1= 294 K
V2= ?
T2= 305 K

Replacing:

$\frac{8.5 L}{294 K}=\frac{V2}{305 K}$

Solving:

$V2=305 K*\frac{8.5 L}{294 K}$

V2=8.8 L

<u><em>The new volume of the balloon is 8.8 L</em></u>

7 0

2 years ago

P7.16 A thin flat plate 55 by 110 cm is immersed in a 6-m/s stream of SAE 10 oil at 20C. Compute the total friction drag if the

timofeeve [1]

Answer:

a

The total friction drag for the long side of the plate is 107 N

b

The total friction drag for the long side of the plate is 151.4 N

Explanation:

The first question is to obtain the friction drag when the fluid i parallel to the long side of the plate

The block representation of the this problem is shown on the first uploaded image

Where the U is the initial velocity = 6 m/s

So the equation we will be working with is

$F = \frac{1}{2} \rho C_fAU^2$

Where $\rho$ is the density of SAE 10W = $870\ kg/m^3$ This is obtained from the table of density at 20° C

$C_f$ is the friction drag coefficient

This coefficient is dependent on the Reynolds number if the Reynolds number is less than $5*10^5$ then the flow is of laminar type and

$C_f$ = $\frac{1.328}{\sqrt{Re} }$

But if the Reynolds number is greater than $5*10^5$ the flow would be of Turbulent type and

$C_f = \frac{0.074}{Re_E^{0.2}}$

Where Re is the Reynolds number

To obtain the Reynolds number

$Re = \frac{\rho UL}{\mu}$

where L is the length of the long side = 110 cm = 1.1 m

and $\mu$ is the Dynamic viscosity of SAE 10W oil $= 1.04*10^{-1} kg /m.s$

This is gotten from the table of Dynamic viscosity of oil

So

$Re = \frac{870 *6*1.1}{1.04*10^{-1}}$

$= 55211.54$

Since 55211.54 < $5.0*10^5$

Hence

$C_f = \frac{1.328}{\sqrt{55211.54} }$

$= 0.00565$

$A$ is the area of the plate = $\frac{ (110cm)(55cm)}{10000}$ = $0.55m^2$

Since the area is immersed totally it should be multiplied by 2 i.e the bottom face and the top face are both immersed in the fluid

$F = \frac{1}{2} \rho C_f(2A)U^2$

$F =\frac{1}{2} *870 *0.00565*(2*0.55)*6^2$

$F = 107N$

Considering the short side

To obtain the Reynolds number

$Re = \frac{\rho U b}{\mu}$

Here b is the short side

$Re =\frac{870*6*0,55}{1.04*10^{-1}}$

$=27606$

Since the value obtained is not greater than $5*10^5$ then the flow is laminar

And

$C_f = \frac{1.328}{\sqrt{Re} }$

$= \frac{1.328}{\sqrt{27606} }$

$= 0.00799$

The next thing to do is to obtain the total friction drag

$F = \frac{1}{2} \rho C_f(2A)U^2$

Substituting values

$F = \frac{1}{2} * 870 * 0.00799 * 2( 0.55) * 6^2$

$= 151.4 N$

4 0

2 years ago

A car travels 10 miles East in 30 minutes. What is its velocity in miles per hour?

Stella [2.4K]

Answer:

Velocity = 20 mph

Explanation:

Distance = 10 miles

Time = 30 min = 0.5 hour

Velocity = distance/time

= 10miles/0.5 hour

= 20 mph

8 0

2 years ago

Read 2 more answers

A centrifuge rotor rotating at 9700 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.96 m

Mice21 [21]

We start from the definition of Torque,

$T = I \alpha$

Where ,

I = moment of inertia

$\alpha =$ Angular acceleration.

The torque given in the problem is 1.96mN.

We look for the moment of inertia of a solid cylinder,

$I = \frac {1} {2} mR ^ 2$

Where m is the mass of 4Kg and R the radius 0.035m

$I = \frac {1} {2} (4) (0.035) ^ 2$

$I = 2.45 * 10 ^ - 3 Kgm ^ 2$

Replacing,

$-1.96 = 2.45 * 10 ^{-3} \alpha \\\alpha = -800rad / s ^ 2$

A) With angular acceleration we can find the number of revolutions, the given equation would be,

$w_f ^ 2-w_i ^ 2 = 2 \alpha \theta$

$0 ^ 2- 9700rpm (2 \pi / 60rpm) ^ 2 = -2 * 800 \ theta$

$\theta = \frac {1031812.3} {1600}$

$\theta = 644.88 revolutions.$

B) We apply the rotational dynamics formula and we can find the time,

$w_f = w_i + \alpha t$

$0 = 9700 rpm (2 \pi / 60 rpm) -800t$

$t = \frac {1015.78} {800}$

$t = 1.26s$

4 0

2 years ago