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Viefleur [7K]
3 years ago
9

heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.5-m-long rope. The ball is pulled to one sid

e and released to swing as a pendulum, reaching a speed of 5.8 m/s as it passes through the lowest point. What is the tension in the rope at that point?

Engineering
1 answer:
aniked [119]3 years ago
8 0

Answer:

Tension on rope = 35.7N

Explanation:

Detailed explanation and calculation is shown in the image

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To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records thr
LenaWriter [7]

This question is incomplete, the complete question is;

To convert from the U.S. Customary (FPS) system of units to the SI system of units. A first-year engineering student records three separate measurements as 653 lb, 69.0 mi/h, and 293 × 10⁶ ft². Suppose this engineering student has to turn in the results, but the professor only accepts results given in SI units.  

Required:

What is the area measurement, 293 × 10⁶ ft², in SI units?

293 × 10⁶ ft² = ?km²

Answer:

the area measurement is  27.221 km²

Explanation:

Given the data in the question;

What is the area measurement, 293 × 10⁶ ft², in SI units

we are to the result of the measured area from ft² to km²

we know that;

1 meter = 3.2808 ft

1 km = 1000 m

1 ft = (1 / 3.2808)m

1 m = ( 1/1000 ) km

since our measured are is 293 × 10⁶ ft²

hence

A = 293 × 10⁶ × [ (1 / 3.2808)m ]²

A = 27221252.74 m²

A = 27221252.74 × [ ( 1/1000 ) km ]²

A = 27.221 km²

Therefore, the area measurement is  27.221 km²

7 0
3 years ago
Convert 86.1 cm to inches​
vaieri [72.5K]
It’ll be, 33.898 inches
6 0
3 years ago
Free brainlist because im new and i just want to but you have t friend me first
Amiraneli [1.4K]
Okay sure.









I’ll 1)chords
2)pulse
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4) the answer is C
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Pretty sure those are the answers
4 0
3 years ago
Jim starts walking from a bus stop. He walks north for 10 km. Next, he walks east for 5 km. He then walks south for 10 km. Next,
Luda [366]

Answer:

b

Explanation:

8 0
3 years ago
A scale model is 4th the size of the pump. Determine the power ratio of the pump and its scale model if the ratio of the heads i
vredina [299]

Given:

size of scale model = 4(size of pump)

power ratio of pump and scale model = 5:1

Solution:

Let the diameter of scale model and pump be d_{s} and d_{p} respectively

and head be  H_{s} and  H_{p} respectively

Now, power, P is given as a function of head(H) and dischagre(Q)

P = \rho gQH                  (1)

From eqn (1):

P \propto QH

and

QH \propto \sqrt{H}D^{2}

So,

P \propto H^{\frac{3}{2}} D^{2}

Therefore,

\frac{P_{s}}{P_{p}} = \frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}

\frac{P_{s}}{P_{p}} = \frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}

\frac{P_{s}}{P_{p}} = \frac{1^{2}\times 5^{\frac{3}{2}}}{4^{2}\times 1^{\frac{3}{2}}}

\frac{P_{s}}{P_{p}} = \frac{5\sqrt{5}}{16}

{P_{s}}:{P_{p}} = {5\sqrt{5}}:{16}

8 0
3 years ago
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