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Anna71 [15]
3 years ago
5

T or F :)

Physics
1 answer:
balandron [24]3 years ago
4 0

That's true.

And I'll go ya a better one:

If the object is moving or not moving, at a constant or changing speed, in a straight or curvy line, and the forces on it do not cancel out and add up to zero, the object will accelerate.

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A circular hole in an aluminum plate is 2.739 cm in diameter at 0.000°C. What is the change in its diameter when the temperature
prisoha [69]

Answer:

L = 2.746 cm

Explanation:

As we know that thermal expansion coefficient of aluminium is given as

\alpha = 24 \times 10^{-6} per ^oC

now we also know that after thermal expansion the final length is given as

L = L_o(1 + \alpha \Delta T)

here we know that

L_o = 2.739 cm

\alpha = 24 \times 10^{-6}

\Delta T = 108 - 0= 108^oC

now we will have

L = 2.739(1 + 24 \times 10^{-6} (108))

L = 2.746 cm

3 0
3 years ago
Which, if any, of the heat engines violate(s) the second law of thermodynamics?
Anuta_ua [19.1K]
The second law states that the total entropy can never decrese over time for an isolated system
3 0
3 years ago
Please help and I don't mean to sound rude but, ONLY ANSWER IF YOUR GOING TO DO ALL 4 QUESTIONS
anzhelika [568]

Answer:

for first question is 2

for second question 1

for third question 2

for forth question 1

Explanation:

i hope i helped

6 0
2 years ago
A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?
seraphim [82]

Answer:

L = 2.8 cm

Explanation:

Period T = 4 / 12 = 1/3 s

T = 2π√(L/g)

L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

6 0
2 years ago
A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched b
Firlakuza [10]

From the given information in the question, the correct option is Option 1: 14 cm.

A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = \frac{1}{2} k x^{2}

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = \frac{1}{2} k x^{2}

10 = \frac{1}{2} k 10^{2}

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = \frac{1}{2} k x^{2}

20 = \frac{1}{2} (0.2) x^{2}

40 = 0.2 x^{2}

x^{2} = 200

x = \sqrt{200}

  = 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

For more information, check at: brainly.com/question/1352053.

8 0
3 years ago
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