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Fiesta28 [93]
2 years ago
13

HELP ME!!! NO LINKS!!! Compare and Contrast the scientific definitions of theory and law. Explain why the first law of thermodyn

amics is not considered to be a theory.
Physics
1 answer:
Nataly_w [17]2 years ago
3 0
The date the model was published.

The use of “laws” originated prior to science splitting from natural philosophy. There’s an implicit assumption that God as the creator laid down both moral and natural laws, with the theologian concerned with the former and the natural philosopher concerned with the latter.

“Theory” begins to take hold in the late 1700s and, very roughly speaking, is used to describe more complex models. “Law” eventually became nearly archaic, although still used to describe very pithy models (Amdahl’s Law, Gustafson’s Law).

The word “model” is gradually superseding both of them.

People have tried to come up with hard-and-fast rules to distinguish them, but scientists are unruly beasts, and use whatever language suits them in the moment.
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Your brother is insisting that you’ll never hear a sound produced behind a barrier wall at the end of your yard you notice that
Tresset [83]

Answer

D.Diffraction

Explanation

Diffraction is a property that is experienced by waves when they come across a barrier when they are in motion.

The ways tends to curve behind the barrier. This is called diffraction of waves.

Now, sound is a wave and it also experience diffraction. . So the brother will be able to hear the sound due to diffraction

8 0
3 years ago
Read 2 more answers
A car traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?
Alenkasestr [34]

accn = - 26m/s divided by 13 secs

accn = 2m/s/s

8 0
3 years ago
According to the theory of plate tectonics, which forces cause the movement of plates in the earth's crust?
Sauron [17]
The correct answer is B.
4 0
3 years ago
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If an object is rolling without slipping, how does its linear speed compare to its rotational speed?
RoseWind [281]

Answer:

v = rw

Explanation:

When an object is rolling continuously without slipping, then every angle it rotates through, is equal to a distance the perimeter has rotated.

If the object completes 10 revolutions and takes a particular time, let's say t to complete it. The angular distance would then be 20 π rad, while its angular velocity will be 20 π/t

The circumference will somehow translate to the distance it covers, which is 20πr, this means that the speed is 20πr/t

So, like the question asked, the linear speed compared to angular speed is

v : w

20πr/t : 20πt, which can be simplified to

r : 1

In essence, v = rw

3 0
3 years ago
a ball kicked with a velocity of 8m/s at an angle of 30 degree to horizontal. calculate the time of flight of the ball. (g=10ms^
posledela

Answer:

Approximately 0.8\; \rm s (assuming that air resistance is negligible.)

Explanation:

Let v_0 denote the initial velocity of this ball. Let \theta denote the angle of elevation of that velocity.

The initial velocity of this ball could be decomposed into two parts:

  • Initial vertical velocity: v_0(\text{vertical}) = v_0 \cdot \sin(\theta).
  • Initial horizontal velocity: v_0(\text{vertical}) = v_0 \cdot \cos(\theta).

If air resistance on this ball is negligible, v_0(\text{vertical}) alone would be sufficient for finding the time of flight of this ball.

Calculate v_0(\text{vertical}) given that v_0 = 8 \; \rm m \cdot s^{-1} and \theta = 30^\circ:

\begin{aligned}& v_0(\text{vertical}) \\ &= v_0 \cdot \sin(\theta) \\ &= \left(8 \; \rm m \cdot s^{-1} \right) \cdot \sin\left(30^{\circ}\right) \\ &= 4\;\rm m \cdot s^{-1} \end{aligned}.

Assume that air resistance on this ball is zero. Right before the ball hits the ground, the vertical velocity of this ball would be exactly the opposite of the value when the ball was launched.

Since v_0(\text{vertical}) = 4\; \rm m \cdot s^{-1}, the vertical velocity of this ball right before landing would be v_1(\text{vertical}) = -4\; \rm m \cdot s^{-1}.

Calculate the change to the vertical velocity of this ball:

\begin{aligned}& \Delta v(\text{vertical}) \\ & = v_1(\text{vertical}) - v_0(\text{vertical}) \\ &= -8\; \rm m \cdot s^{-1}\end{aligned}.

In other words, the vertical velocity of this ball should have change by 8\; \rm m \cdot s^{-1} during the entire flight (from the launch to the landing.)

The question states that the gravitational field strength on this ball is g = 10\; \rm m \cdot s^{-2}. In other words, the (vertical) downward gravitational pull on this ball could change the vertical velocity of the ball by 10\; \rm m\cdot s^{-1} each second. What fraction of a second would it take to change the vertical velocity of this ball by 8\; \rm m \cdot s^{-1}?

\begin{aligned}t &= \frac{\Delta v(\text{initial})}{g} \\ &= \frac{8\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-2}} = 0.8\; \rm s\end{aligned}.

In other words, it would take 0.8\; \rm s to change the velocity of this ball from the initial velocity at launch to the final velocity at landing. Therefore, the time of the flight of this ball would be 0.8\; \rm s\!.

5 0
3 years ago
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