PHYSICS HIGH SCHOOL

A net force of 40 N south acts as an object with a mass of 20kg. What is the object's acceleration

Answers

Answer 1

Answer:

Answer:

a = 2 m/s²

Explanation:

Given: 20 g, 40N

To find: Acceleration (a)

Solution: To find the acceleration (A), divide the force by the weight

A = F ÷ m

= 40 ÷ 20

= 2 m/s²

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

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A scientist told his student assistants that he had sorted groups of mice by their intelligence and asked the assistants to assess the groups’ maze-solving abilities. The scientist had not actually sorted the mice by intelligence. The students rated the “highest intelligence” group of mice as the best maze solvers and the “lowest intelligence” group as the worst maze solvers. What was the scientist trying to demonstrate by doing this experiment?Experimental subjects should never be put into groups. Mice should not be judged by their intelligence. Higher intelligence mice are better maze solvers. People can introduce their own biases into an experiment.

A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of begin quantity omega times t end quantity, where V is the maximum voltage, omega is the angular frequency, and t is the time. The current supplied by this source that flows through this resistor is described with the function i of t is equal to I times cosine of begin quantity omega times t end quantity, where I is the maximum current. What is the average power supplied by this AC source?

Answers

Answer:

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

$p(t)=v(t)i(t)$

So the average power is:

$P=(1)/(T)\intop_(0)^(T)p(t)dt$

But:

$v(t)=v_(m)cos(\omega t) \n \n i(t)=i_(m)cos(\omega t)$

So:

$P=(1)/(T)\intop_(0)^(T)v_(m)cos(\omega t)i_(m)cos(\omega t)dt \n \n P=(v_(m)i_(m))/(T)\intop_(0)^(T)cos^(2)(\omega t)dt \n \n But: cos^(2)(\omega t)=(1+cos(2\omega t))/(2)$

$P=(v_(m)i_(m))/(T)\intop_(0)^(T)((1+cos(2\omega t))/(2) )dt \n\nP=(v_(m)i_(m))/(T)\intop_(0)^(T)[(1)/(2)+(cos(2\omega t))/(2)]dt \n\nP=(v_(m)i_(m))/(T)[(1)/(2)(t)\right|_0^T +(sin(2\omega t))/(4\omega) \right|_0^T] \n \n P=(v_(m)i_(m))/(2T)[(t)\right|_0^T +(sin(2\omega t))/(2\omega) \right|_0^T] \n \n P=(v_(m)i_(m))/(2)$

In terms of RMS values:

$V_(RMS)=V=(v_(m))/(√(2)) \n \n I_(RMS)=I=(i_(m))/(√(2)) \n \n Then: \n \n P=VI$

A gas has an initial volume of 2.5 L at a temperature of 275 K and a pressure of 2.1 atm. The pressure of the gas increases to 2.7 atm, and the temperature of the gas increases to 298 K.The final volume of the gas, rounded to the nearest tenth, is.

Answers

The following information are given:
V1 = 2.5L
T1 = 275K
P1 = 2.1 atm
T2 = 298K
P2 = 2.7 atm
V2 = ?
The formular for solving the equation is: P1V1T1 = P2V2T2
V2 = P1V1T1/P2VT2
V2 = 2.5 * 275 * 2.7 * 298 = 1443.75/804.6 = 1.794
Therefore, V2 = 1.8L.

Its 2.1, I just took the test, the verified answer is wrong.

Which atomic number is the threshold value below which fusion may occur?

Answers

Answer:

The atomic number 26(iron) is the threshold value below which the fusion might occur.

Explanation:

Nuclear fusion is a reaction in which two or more nuclei are combined to form one or more different atomic nuclei and subatomic particles.

Energy released in a fusion reaction is because of a key feature of nuclear matter called the binding energy which is a measure of the efficiency with which its constituent nucleons are bound together.

As we go up in atomic number, the energy released per nuclei goes down until it hits a minimum which is for atomic number 26 (iron) and fusion is not possible.

Answer:

Atomic no. 26 i.e. Iron

Explanation:

Nuclear Fusion is a process in which two or more nucleus are fused(combined) together to make heavier nucleus and release a lot of energy and other subatomic particles in the process. Considerable amount of energy is required to fuse the nucleus together but the output energy is much more than the input energy. It becomes an exothermic process.

In case of heavier nuclei, this doesn't hold true as the process becomes endothermic. That means the output energy will be way lesser than the input energy given to fuse the nucleus. This happens for elements of atomic no. 26 or above. Even in stars we can find element till atomic no. 26 as they can be formed by fusion. When a star reaches to iron fusion stage that is the end of it and it explodes into a supernova.

AP PHYSICS please answer

Answers

I think it’s d not sure

A growing person needs a diet that supplies about 10MJ of energy per day. Calculate the amount of energy supplied by such a diet each second, and hence the person’s average power. (Give your answer to the nearest 10W)

Answers

We are given that a growing person needs a diet that supplies about 10 MJ of energy per day.

To convert this to energy supplied per second, we can use the following conversion:

1 day = 24 hours/day x 60 minutes/hour x 60 seconds/minute = 86400 seconds

Therefore, the energy supplied per second is:

10 MJ/day = 10,000,000 J/day ÷ 86400 seconds/day = 115.74 J/s

The person's average power is the energy supplied per second, which is:

P = 115.74 J/s ≈ 120 W (rounded to the nearest 10 W)

Therefore, the person's average power is approximately 120 W.

To know more about average power, click here:-

brainly.com/question/30888338

#SPJ11

Visible light is an example of?

Answers

Answer:

Visible light is an example of electromagnetic waves.

Explanation:

Visible light is an example of electromagnetic waves. The visible light are the range in which the light is visible to the human eye. The wavelength of visible light varies from 400 nm to 700 nm.

The visible region consists of following light namely,

1. Violet : 380–450 nm

2. Blue : 450–485 nm

3. Cyan : 485–500 nm

4. Green : 500–565 nm

5. Yellow : 565–590 nm

6. Orange : 590–625 nm

7. Red : 625–740 nm

Hence, the visible light is an example of electromagnetic wave.

electromagentic waves

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