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Helen [10]
2 years ago
15

In a particular application involving airflow over a surface, the boundary layer temperature distribution may be approximated as

T−Ts/T[infinity]−Ts=1−exp⁡(−Pr(u[infinity]y)/????) Where y is the distance normal to the surface and the Prandtl number, Pr=cpμ/k=0.7, is a dimensionless fluid property. If T[infinity]=400K, Ts=300K, and u[infinity]/υ=5000 m-1, what is the surface heat flux? 6.11

Physics
1 answer:
Anni [7]2 years ago
8 0

Answer:

The Surface heat flux is -9205 W/m^2

Explanation:

 Explanation is in the following attachment    

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A motion sensor is used to create the graph of a student’s horizontal velocity as a function of time as the student moves toward
Marrrta [24]

Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

                     Area of trapezoidal are=

                            =\frac{1}{2} *(1.5+0.75)+\frac{1}{2} (1+0.75)(-0.5)

                            =\frac{1}{2} *(2.25-1.75*0.5)

                            =0.6875 m

As the area is positive therefore displacement from xo is positive

                                  ax=(change in velocity)/(Time)

                                  ax=\frac{-0.5-0}{3} =-\frac{1}{6} ms2

3 0
2 years ago
A future use of space stations may be to provide hospitals for severely burned persons. it is very painful for a badly burned pe
natta225 [31]
<span>1.5 minutes per rotation. The formula for centripetal force is A = v^2/r where A = acceleration v = velocity r = radius So let's substitute the known values and solve for v. So F = v^2/r 0.98 m/s^2 = v^2/200 m 196 m^2/s^2 = v^2 14 m/s = v So we need a velocity of 14 m/s. Let's calculate how fast the station needs to spin. Its circumference is 2*pi*r, so C = 2 * 3.14159 * 200 m C = 1256.636 m And we need a velocity of 14 m/s, so 1256.636 m / 14 m/s = 89.75971429 s Rounding to 2 significant digits gives us a rotational period of 90 seconds, or 1.5 minutes.</span>
5 0
2 years ago
A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm.
mote1985 [20]

Answer:

a) When its length is 23 cm, the elastic potential energy of the spring is

0.18 J

b) When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

Explanation:

Hi there!

a) The elastic potential energy (EPE) is calculated using the following equation:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretched lenght.

Let´s calculate the elastic potential energy of the spring when it is stretched 3 cm (0.03 m).

First, let´s convert the spring constant units into N/m:

4 N/cm · 100 cm/m = 400 N/m

EPE = 1/2 · 400 N/m · (0.03 m)²

EPE = 0.18 J

When its length is 23 cm, the elastic potential energy of the spring is 0.18 J

b) Now let´s calculate the elastic potential energy when the spring is stretched 0.06 m:

EPE = 1/2 · 400 N/m · (0.06 m)²

EPE = 0.72 J

When the stretched length doubles, the potential energy increases by a factor of four to 0.72 J

7 0
2 years ago
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
2 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
1 year ago
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