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Ede4ka [16]
2 years ago
12

4: Question 2 - 10.0 pts possible

Physics
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=0.03m^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
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