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saw5 [17]
3 years ago
14

Newton's law of universal gravitation, the law of conservation of mass, the law of superposition, and the laws of thermodynamics

are all examples of scientific laws because..
A.
they have been modified repeatedly over time.
B.
they have been approved and accepted by congress.
C.
there are no exceptions to these laws under specific stated conditions.
D.
they are found in science textbooks.
Physics
1 answer:
Llana [10]3 years ago
5 0

Answer:

D

Explanation:

There are no exceptions to these laws under specific stated conditions..

The laws of science are statements that describe or predict a range of phenomena as they appear in nature. Laws explain facts determined by experiment. Laws do not have absolute certainty and be changed by future experiments and observations.

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9) Of all the types of light the Sun gives off, it emits the greatest amount of light at visible wavelengths of light. If the Su
erastova [34]

Answer:

*  most of the emission would be in the infrared part, the visible radiation would be very small.

*total intensity of the semition decreases that the intensity depends on the fourth power of the temperature

Explanation:

The radiation emitted by the Sun is approximately the radiation of a black body, if the Sun were to cool, the maximum emission wavelength changes

          λ T = 2,898 10⁻³

          λ = 2,898 10⁻³ / T

if the temperature decreases the maximum wavelength the greater values ​​are moved, that is to say towards the infrared. Therefore the emission curve also moves, in this case most of the emission would be in the infrared part, the visible radiation would be very small.

Furthermore, the total intensity of the semition decreases that the intensity depends on the fourth power of the temperature according to Stefan's law

           P = σ A eT⁴

7 0
3 years ago
When the temperature of matter decrease , the particles do what
Rom4ik [11]

Answer:

When the temperature decreases the particals start to slow down.

4 0
3 years ago
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

5 0
3 years ago
Is 250 g higher than 100 g in science
polet [3.4K]
Its the same thing. Is 250 grams more then 100 grams
5 0
3 years ago
Read 2 more answers
How much net force is required to accelerate a 2000 kg car at 3.00 m/s^2
andrezito [222]

The net force required to accelerate a car is 6000 N.

Force is defined as the product of the mass and acceleration of the body. Force is used to changing the velocity that is to accelerate an object or a body of a particular mass. The unit of Force is Newton or kg m/s^2.

The formula used to calculate the net force is :

F = ma

where, F = Force

m = mass = 2000 kg

a = acceleration = 3.00 m/s^2

∴ F = 2000*3

F = 6000 N

Thus, to accelerate the car at 3.00 m/s^2 of mass 2000 kg net force required is 6000 N.

To learn more about force,

brainly.com/question/1046166

6 0
1 year ago
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