Look at jackie's planned workout. Based on her entries, what would she do to maintain an effective long term workout routine

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

3 years ago

Which of Newton’s Laws explains why an object originally moving in a circular path will continue in a straight line path if the

Andreas93 [3]

When object is moving in circular path then if suddenly its force reduced to zero then it will move to straight line

This is defined by Newton's First Law

So as per first law of Newton we can say

When there is no net force on an object then it will continue its state of motion in its original state without any change in it.

This is also known as law of inertia

so correct answer will be

Newton's 1st law

6 0

3 years ago

A spelunker is surveying a cave. She follows a passage 180 m straight west, then 210 m in a direction 45° east of south, and the

Maksim231197 [3]

got this from assignmentexpert

7 0

3 years ago

If you walk 1.4 km north and then 4.8 km east, what are the magnitude and direction of your resultant displacement?

Amiraneli [1.4K]

B - magnitude: 5 km; direction: 73.7° east of north

5 0

3 years ago

A block is placed on an inclined plane with an angle of inclination θ (in degrees) with respect to horizontal. The coefficient o

hjlf

Answer:

The maximum value of θ that will cause the block to remain stationary on the inclined surface is 21.8°

Explanation:

Given;

coefficient of static friction, μ = 0.4

for the block to remain stationary on the inclined plane, force pushing the block upward must be equal to the force acting downwards.

μR = mgsinθ

μmgcosθ = mgsinθ

μcosθ = sinθ

μ = sinθ/cosθ

μ = tanθ

θ = tan⁻¹(0.4) = 21.8°

Therefore, the maximum value of θ that will cause the block to remain stationary on the inclined surface is 21.8°

6 0

3 years ago