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3 years ago

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