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Vadim26 [7]
3 years ago
11

As you increase the number of supporting strings, what happens to the force needed to lift the bottom block

Physics
1 answer:
Alexus [3.1K]3 years ago
8 0
The weight is spread across multiple different strings, allowing for each string to require dealing with a large amount of stress, because the stress is evenly distributed throughout each string. hope i helped :3
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Mass = 1.2 kg = 1200 grams.

Volume = mass/density = 1200 cm3.

Hope this helps!
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The passenger-side rear view mirror on a car says, “Objects in the mirror may be closer than they appear”. Assuming the images a
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Answer:

the mirror is a convex mirror

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The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F
photoshop1234 [79]

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

<em>T</em> - 25 N = (8 kg) <em>a</em>

where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.

The hanging mass has a net force of

(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>

where <em>g</em> = 9.8 m/s².

Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :

(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>

33.8 N = (14 kg) <em>a</em>

<em>a</em> = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

<em>T</em> - 25 N = (8 kg) (2.4 m/s²)

<em>T</em> ≈ 25 N + 19.31 N ≈ 44 N

5 0
3 years ago
Which disease do you think is most easily spread? 5 Answers
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3 years ago
A power supply maintains a potential difference of 52.3 V 52.3 V across a 1570 Ω 1570 Ω resistor. What is the current in the res
irakobra [83]

Answer:

0.033 A

Explanation:

Current: This can be defined as the rate of flow of electric charge in a circuit.

The S.I unit of current is Ampere (A)

From Ohm's law.

V = IR ............................ Equation 1

Where V = Potential difference, I = current, R = resistance.

Making I the subject of the equation,

I = V/R................... Equation 2

Given: V = 52.3 V, R = 1570 Ω

Substitute into equation 2

I = 52.3/1570

I = 0.033 A.

Hence the current in the resistor = 0.033 A

6 0
3 years ago
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