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2 years ago

How are theories and laws connected

Physics

2 answers:

Anni [7]2 years ago

4 0

Answer:

Laws are statements about something that's been observed and stated while a theory is an explanation of what's been observed. This connection between them forms a main idea that many people regulate as "what's normal."

Explanation:

nasty-shy [4]2 years ago

4 0

Answer:

A hypothesis is a limited explanation of a phenomenon; a scientific theory is an in-depth explanation of the observed phenomenon. A law is a statement about an observed phenomenon or a unifying concept, according to Kennesaw State University. ... However, Newton's law doesn't explain what gravity is, or how it works.

for e.g:

The particle theory and Darwin's theory of human origin.

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A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

$P_r = 345 Watt$

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

$\Delta V = E + i r$

here we have

$E = 12 V$

$i = 56 A$

$r = 0.11$

now we have

$\Delta V = 12 + (0.11)(56) = 18.16 V$

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

Part c)

Rate of energy conversion into EMF is given as

$P_{emf} = i E$

$P_{emf} = (56)(12)$

$P_{emf} = 672 Watt$

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

$V = E - i r$

$V = 12 - (56)(0.11)$

$V = 12 - 6.16 = 5.84 V$

Part e)

now the rate of energy dissipation is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

7 0

2 years ago

Differences between freezing point and melting point (Atleast 5 differences)

labwork [276]

Answer:

What is freezing point?

A liquid's freezing point is determined at which it turns into a solid. Corresponding to the melting point, the freezing point often rises with increasing pressure. In the case of combinations and for some organic substances, such as lipids, the freezing point is lower than the melting point. The first solid which develops when a combination freezes often differs in composition from the liquid, and the development of the solid alters the composition of the remaining liquid, typically lowering the freezing point gradually. Utilizing successive melting and freezing to gradually separate the components, this approach is used to purify mixtures.

What is melting point?

The temperature at which a purified substance's solid and liquid phases may coexist in equilibrium is referred to as the melting point. A solid's temperature goes up when heat is added to it until the melting point is achieved. The solid will then turn into a liquid with further heating without changing temperature. Additional heat will raise the temperature of the liquid once all of the solid has melted. It is possible to recognize pure compounds and elements by their distinctive melting temperature, which is a characteristic number.

The difference between freezing point and melting point:

While a substance's melting point develops when it transforms from a solid to a liquid, a substance's freezing point happens when a liquid transforms into a solid when the heat from the substance is removed.
When the temperature rises, the melting point can be seen, and when the temperature falls, the freezing point can be seen.
When a solid reaches its melting point, its volume increases; meanwhile, when a liquid reaches its freezing point, its volume decreases.
While a substance's freezing point is not thought of as a distinctive attribute, its melting point is.
While external pressure is a significant component in freezing point, atmospheric pressure is a significant element in melting point.
Heat must be supplied from an outside source in order to reach the melting point for such a state shift. When a material is at its freezing point, heat is needed to remove it from the substance in order to alter its condition.

<em>Reference: Berry, R. Stephen. "When the melting and freezing points are not the same." Scientific American 263.2 (1990): 68-75.</em>

7 0

1 year ago

A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

7 0

3 years ago

A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c

Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

$I= \Delta p = m \Delta v = m (v-u)$

where

m is the mass of the object

$\Delta v$ is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

$I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s$

(b) 155 N

The impulse can also be rewritten as

$I=F \Delta t$

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

$\Delta t$ is the duration of the collision

In this situation, we have

$\Delta t = 0.06 s$

So we can re-arrange the equation to find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N$

6 0

3 years ago

What happens to air pressure as altitude decreases

Ksivusya [100]

Pressure decreases with increasing altitude. The pressure at any level in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels.

8 0

3 years ago

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How are theories and laws connected ​

How are theories and laws connected