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3 years ago

What did Greek philosophers, Isaac Newton, and others commonly use to explain

Physics

1 answer:

Temka [501]3 years ago

3 0

Answer:

Hey there!

They used math to explain their observations, for example, Newton is famous for his three laws, and universal laws of gravitation.

Let me know if this helps :)

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Potential energy of an apple is 6j. the apple is 3.00m high. what is the mass of the apple?

HACTEHA [7]

Answer:

0.2 kg

Explanation:

PE = mgh

6 J = m (9.8 m/s²) (3.00 m)

m = 0.2 kg

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3 years ago

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Amanda [17]

Answer:

decantation, distilling, freezing

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3 years ago

A 15.0-kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0° above the horizontal. The block is

Tatiana [17]

Answer:

Explanation:

from the question we are told that

Load $L=15kg$

Force $F=70N$

Angle of inclination $=20.0 degres$

Displacement $m=5 meters$

coefficient of kinetic friction $\alpha =0.300$

3 0

3 years ago

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with

Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

$\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.$

Writing for each direction at a time,

$m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.$

Their kinetic energy changed by

$K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) = \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2 + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.$