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Katena32 [7]
3 years ago
11

Marble A is placed in a hollow tube, and the tube is swung in a horizontal plane causing the marble to be thrown out. As viewed

from the top, which of the following choices best describes the path of the marble after leaving the tube?

Engineering
1 answer:
Vilka [71]3 years ago
6 0

Answer:

<em>The direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

Explanation:

The question is incomplete. So the picture, which is missing in question, is attached for your review.

As it can be seen in the picture, the ball coming out of the tube will have two components of velocity. One is along the length of tube (because ball is moving in that direction and is coming out from the hole), other is velocity component will be perpendicular to the tube (because the ball is made to move in that direction as the tube is rolling on the surface).

<em>So, taking the resultant of two vectors of velocity, the resultant direction of ball will be Number 4 (as can be seen in attached picture) ---- the path of ball will be making some angle when it leaves the tube. </em>

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1. A screw driver with a 1.5 inch diameter handle is used to install a 1/4-20 UNC screw. Determine the circumference where the e
marishachu [46]

Explanation:

1. Circumference of the handle (where the effort is applied) is pi times diameter.

C = πd

C = 1.5π inches

C ≈ 4.71 inches

2. Mechanical advantage is the ratio of distance in over distance out.

MA = din / dout

MA = 1.5 / 0.25

MA = 6

3. Mechanical advantage is the ratio of force out over force in.

MA = Fout / Fin

6 = Fout / 10 lb

Fout = 60 lb

8 0
3 years ago
A rope having a weight per unit length of 0.4 lb/ft is wound 2 1/2 times around a horizontal rod. Knowing that the coefficient o
Mademuasel [1]

Explanation:

Tension in the rope

\begin{aligned}T_{1} &=0.4 \times x \\T_{2} &=100+0.4 \times 10 \\&=104\end{aligned}

M s=0.3  ∅  =2 \cdot 5(2 \pi)=5 \pi

\begin{aligned}\frac{T_{1}}{T_{2}}=e^{\mu \theta} & \Rightarrow \frac{104}{0.4 x}=e^{0.3(5 \pi)} \\& \Rightarrow x=2.34 \mathrm{H}\end{aligned}

NOTE : Refer the image

3 0
3 years ago
___________ is NOT a common injury that an automotive tech may experience at work.
Degger [83]

Answer:The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

Explanation:

The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

7 0
3 years ago
A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu
bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

a.

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

b.

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

7 0
4 years ago
0/5 pts
Brilliant_brown [7]

Explanation:

150 divide by 150 and that how you do the is you what to divide together 15/ 150 you welcome have a good day is you need something else

4 0
3 years ago
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