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3 years ago

9

A box of mass 10 kg is pulled from the hold of a ship with an acceleration of 1 m/s^2 by a vertical rope attached to it .find th

e tension in the rope.

1 answer:

wlad13 [49]3 years ago

5 0

F=ma
Mass times acceleration
We have g (10ms^_2) and a (1 given)
So total would be
10 kg times (10+1) =
110 N

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The best and most correct answer among the choices provided by your question is the third choice or letter C.

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Read 2 more answers

A motorcyclist accelerates from rest to 10 mi/hr. what is the change in velocity

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The change in velocity is 10 mi/h (4.47 m/s)

Explanation:

The change in velocity of the motorcyclist is given by

$\Delta v = v-u$

where

v is the final velocity

u is the initial velocity

In this problem, we have

u = 0 (the motorbike starts from rest)

v = 10 mi/h

Therefore, the change in velocity is

$\Delta v = 10 -0 = 10 mi/h$

And keeping in mind that

1 mile = 1609 m

1 h = 3600 s

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$\Delta v = 10 \frac{mi}{h} \cdot \frac{1609 m/mi}{3600 s/h}=4.47 m/s$

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3 years ago

An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is giv

konstantin123 [22]

Answer:

8.91 %

Explanation:

Since v² = 2gy

By the relative error formula,

2Δv/v = Δg/g + Δy/y multiplying by 100%, we have

2Δv/v × 100% = Δg/g × 100 % + Δy/y × 100%

2(Δv/v × 100%) = Δg/g × 100 % + Δy/y × 100%

Δg/g × 100 % = 2(Δv/v × 100%) - Δy/y × 100%

Since Δv/v × 100% = 3.69 % and Δy/y × 100% = 5 %

Since we have a difference for the percentage error in g, we square the percentage errors and add them together. So,

[Δg/g × 100 %]² = [2(Δv/v × 100%)]² + [Δy/y × 100%]²

[Δg/g × 100 %]² = [2(3.69)]² + [5%]²

[Δg/g × 100 %]² = [4)(3.69 %)² + [5%]²

[Δg/g × 100 %]² = 54.4644 %² + 25%²

[Δg/g × 100 %]² = 79.4644 %²

taking square-root of both sides, we have

[Δg/g × 100 %] = 8.91 %

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3 years ago

We start with 5.00 moles of an ideal monatomic gas with an initial temperature of 128 ∘C. The gas expands and, in the process, a

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Answer:

The final temperature of the gas is <em>114.53°C</em>.

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Secondly, from the ideal gas law, we calculate the final temperature of the gas, using the change in internal energy:

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Therefore the final temperature of the gas, T₂, is 114.53°C.

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