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liraira [26]
4 years ago
8

In this experiment, you will use a track, a toy car, and some washers to explore Newton’s first two laws of motion. You will mak

e observations and collect data regarding the motion of these objects. In the space below, write a general scientific question that you will answer by doing this experiment.
Physics
2 answers:
polet [3.4K]4 years ago
7 0

How can we experimentally verify newton's laws?

borishaifa [10]4 years ago
3 0

Answer:

How can Newton's laws be experimentally verified?

Explanation:

Just took it right now

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During which two phases would surfer’s most likely benefit
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explain how you can determine if a chemical reaction a single replacement reaction or a double replacement reaction
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8 0
3 years ago
There are 640 acres in a square mile. how many square meters are there in 7.80 acres.
lukranit [14]

Answer:

On 7.8 acres there are 31565.3 square meters

7.8 acres = 31565.3 m²

Explanation:

Conceptual analysis

Acre, square mile and square meter are area units, so their equivalences are in square units.

Known data

1 (mi)² = 640 acres

1 mi = 1609.34 m

Problem development

We know that 1 mile = 1609.34m, then: (1 mile)² = (1609.34m)²=(1609.34)²m²

We apply the known conversion factors to calculate the square meters (X) in 7.8 acres:

X=7.8 acres *\frac{(1 mi)^{2} }{640 acres} * \frac{(1609.34)^{2}m^{2}  }{(1 mi)^{2} }

We eliminate acres and square miles (mi²) because they are in the numerator and denominator, so the answer is in square meters (m²)

X=31565.3 m²

On 7.8 acres there are 31565.3 square meters

7.8 acres = 31565.3 m²

8 0
3 years ago
A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab
Drupady [299]

Answer

According the conservation of energy

\dfrac{1}{2}mv_i^2+\dfrac{1}{2}I\omega_i^2+0 = mgh + 0

I for ball = \dfrac{2}{3}mr^2

\dfrac{1}{2}mv_i^2+\dfrac{1}{2} \dfrac{2}{3}mr^2\omega_i^2= mgh

\omega_i = \dfrac{v_i}{r}

v_i^2+\dfrac{2}{3}r^2(\dfrac{v_i}{r})^2 = 2gh

v_i^2+\dfrac{2}{3}v_i^2 = 2gh

v_i^2+[1+\dfrac{2}{3}]=2gh

v_i^2\dfrac{5}{3}=2gh

v_i=\sqrt{\dfrac{6gh}{5}}=\sqrt{\dfrac{6\times 9.8\times 5}{5}}

v_i = 7.67\ m/s

a) \omega_i = \dfrac{v_i}{R}

\omega_i = \dfrac{7.67}{0.113}

\omega_i =67.87\ rad/s

b) K_{rot} = \dfrac{1}{2}\dfrac{2}{3}mR^2\omega_i^2

   K_{rot} = \dfrac{1}{3}\times 0.426\times 0.112^2\times 67.87^2

   K_{rot} = 8.205\ J

4 0
3 years ago
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