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2 years ago

12

4: Question 2 - 10.0 pts possible

1 answer:

Orlov [11]2 years ago

8 0

Answer:

5. All of the statements are true; non is false

Explanation:

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If the force squeezing two surfaces together is decreased, the force of dry sliding friction between the two surfaces will most

Anastasy [175]

<em>If the force squeezing two surfaces together is decreased, the force of dry sliding friction between the two surfaces will most likely decrease. </em>

<u>therefore your answer is B)</u><u>d</u><u>e</u><u>crease </u>

Hope this helps you- have a good day bro cya)

7 0

2 years ago

A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent

solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

Kinetic Energy K.E

Solution

From the moment of inertia

$I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}$

The angular acceleration is given as

$a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}$

The angular velocity is given as

$w=at\\w=(0.905)(2.55)\\w=2.31rad/s$

So the Kinetic Energy is given as

$K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J$

3 0

3 years ago

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

4. A car accelerates uniformly from rest at 3.2 m/s^.

Oksi-84 [34.3K]

Answer:

A

Explanation:

7 0

3 years ago

After driving a portion of the route, the taptap is fully loaded with a total of 26 people including the driver, with an average

Svet_ta [14]

Answer:

0.44807175m

Explanation:

k = Spring constant = $4\times 10^{4}\ N/m$ (Assumed, as it is not given)

g = Acceleration due to gravity = 9.81 m/s²

Total mass is

$m=26\times 67+3\times 15+5\times 3+25\\\Rightarrow m=1827\ kg$

The total weight will balance the spring force

$mg=kx\\\Rightarrow x=\dfrac{mg}{k}\\\Rightarrow x=\dfrac{1827\times 9.81}{4\times 10^{4}}\\\Rightarrow x=0.44807175\ m$

The springs are compressed by 0.44807175m

5 0

3 years ago