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Firdavs [7]
3 years ago
9

For which of these questions could a testable hypothesis be developed? Check all that apply.

Physics
2 answers:
Dima020 [189]3 years ago
8 0
-- <span>Does the width of a rubber band affect how far it will stretch?

-- How does the thickness of a material affect insulation?

You could take a guess at the answer to each of these,
and then test to see if your guess is correct.


</span>
timurjin [86]3 years ago
7 0

Answer:

A.) Does the width of a rubber band affect how far it will stretch?

B.) How does the thickness of a material affect insulation?

C.) Do all objects fall to the ground at the same speed?

Explanation:

just did the assignment

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The periodic table arranges elements by increasing _________.
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Atomic number

Explanation:

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5. Two charged particles are separated by a distance of 12 meters. The Coulomb force between them is 20 N. What will the Coulomb
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Answer:

A) 80 N

Explanation:

The closer the particles get, the stronger the Coulomb force, which elongates choices C and D. The Coulomb force is inversely proportional to the distance squared. If the distance is cut in half, the force is multiplied by the reciprocal of (1/2)^2, which is 4. Multiplying it out, 20 times 4 is 80 N.

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What happens when the sun emits more energy than normal
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4 0
3 years ago
A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

4 0
3 years ago
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