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3 years ago

Why are peer reviews important?

Physics

2 answers:

Pavlova-9 [17]3 years ago

7 0

Answer:

Peer reviews are important because individuals with similar background knowledge are used to review work to detect inaccuracies or deficiencies in the work.

Explanation:

Those individuals with similar educational and work background knowledge can more accurately review work since they have the required knowledge base for the subject.

kaheart [24]3 years ago

3 0

Answer and explanation;

Peer review is a very important process that authors have to go through before they are able to publish their research manuscript. The main purpose is to ensure that whatever comes out published is in its excellent form, i. e., virtually free of errors.

Benefits;

1, Corrects vague terms
2, Provides feedback as to the effectiveness of your communication
3, Allows you to see other people’s perspectives on issues raised
4, Prevents you from committing serious blunders in your arguments
5, Gives confidence
6, Facilitates concise writing
7, Improves Grammar
8, Allows you to expound on your points
9, Confirms your observations
10, Encourages you to perform better next time

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A force of 1000 newtons was necessary to lift a rock. A total of 3000 joules of work was done. How far was the rock lifted?

zhenek [66]

Answer: 3 meters

Explanation:

If you were to use a 3 line answer then it would look like this:

d = W/F

d = 3000/1000

d = 3 meters

8 0

2 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,

valina [46]

Answer:

$=2.99\times10^9J$

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement $S = 0.72\times10^3m$

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

= 2×F×cos 19° ................. (1

therefore , total work is

W = F'S

= (2F cos19)×S

$= (2)(2.20\times10^6 N)cos19° (0.720\times10^3 m)$

$=2.99\times10^9J$

6 0

3 years ago

A billiard ball travels 23 cm in the positive direction , hits the cushion and rebounds in the negative direction , and Finally

wel

Explanation :

Displacement refers to the distance between the final and the initial position. Hence the displacement of the ball will be the difference between the initial and the final displacement.

Let the initial position be 0.

Final position = 8 cm

So the difference between initial position and final position = 0 – 8 = - 8 cm.

So the billiard ball comes to rest 8.0 cm behind its orbital position.

4 0

3 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

igomit [66]

Answer:

The maximum electric power output is $P_{max} =1.339*10^{9} \ W$

Explanation:

From the question we are told that

The capacity of the hydroelectric plant is $\frac{V}{t} = 690 \ m^3 /s$

The level at which water is been released is $h = 220 \ m$

The efficiency is $\eta =$ 0.90

The electric power output is mathematically represented as

$P = \frac{PE_l - PE _o}{t}$

Where $PE_l$ is the potential energy at level h which is mathematically evaluated as

$PE_l = mgh$

and $PE_o$ is the potential energy at ground level which is mathematically evaluated as

$PE_o = mg(0)$

$PE_o = 0$

$P = \frac{mgh}{t}$

here $m = V * \rho$

where V is volume and $\rho$ is density of water whose value is $\rho = 1000 kg/m^3$

$P = \frac{(\rho * V) * gh}{t}$

$P = \frac{V}{t} * gh \rho$

substituting values

$P =690 * 9.8 * 220 * 1000$

$P =1.488*10^{9} \ W$

The maximum possible electric power output is

$P_{max} = P * \eta$

substituting values

$P_{max} =1.488*10^{9} * 0.90$

$P_{max} =1.339*10^{9} \ W$

6 0

3 years ago

One normal afternoon in undisclosed City X, the chocolate factory workers overload the Easter egg machine. A group of angsty tee

musickatia [10]

Answer:

The correct option is;

C. 1,715 m

Explanation:

We are given the information from the group of teen at the City edge

Time of arrival of explosion sound = 5 s after sighting

Time of sighting explosion = 5 s before hearing the boom

Speed of sound in air ≈ 343 m/s

Speed of light = 299,792 km/s

Therefore, distance covered by sound in 5 seconds is given by the following equation;

$Speed = \frac{Distance}{Time}$

$\therefore 343 \ m/s= \frac{Distance}{5 \, s}$

Hence Distance = 343 m/s × 5 s = 1715 m

To check, we compare the time it would take for the light to cover 1715 m

That is $Time = \frac{Distance}{Speed} = \frac{1715}{299,792,000} = 0.00000572 \, s$ which is instantaneous hence the distance can be approximated by the time duration for the speed of sound.

Therefore, the distance of the students from the factory is approximately 1,715 m

8 0

3 years ago