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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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Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

3 years ago

A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to ro

NARA [144]

Answer:

the required minimum magnitude of the force F is 21 N

Explanation:

Given the data in the question,

m = 5 kg

width = 60 cm

height = 80 cm

Let force is F represent in the image below,

so when the block about to rotate normal shifted to edge of cube

mg(w/2) = Fh

F = mg(w/2) / h

we know that g = 9.8 m/s²

we substitute

F = (5 × 9.8 ( 60/2)) / 70

F = (5 × 9.8 × 30 ) / 70

F = 1470 / 70

F = 21 N

Therefore, the required minimum magnitude of the force F is 21 N

5 0

3 years ago

A particle moves according to the equation x = 11t^2, where x is in meters and t is in seconds.

Savatey [412]

We are given the equation:

x = 11t^2

We use that equation to calculate for the distance traveled.
For (a)

At t=2.20 sec,
x =53.24 meters

At t=2.95 sec,
x =95.73 meters

Velocity = (95.73 meters - 53.24 meters) / (2.95 s - 2.20 s ) = 56.65 m/s

For (b)

At t=2.20 sec,
x =53.24 meters

At t=2.40 sec,
x =63.36 meters

Velocity = (63.36 meters - 53.24 meters) / (2.40 s - 2.20 s ) = 50.6 m/s

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3 years ago

How do you find the speed of an electromagnetic wave? Multiply the wavelength by the frequency. Dvide the frequency by the wavel

Leokris [45]

Answer:

Multiply the wavelength by the frequency.

Explanation:

The velocity of a wave is the frequency times the wavelength.

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3 years ago

Acceleration is defined as the change in velocity divided by

denpristay [2]

Answer:

Time elapsed

Explanation:

Acceleration is a vector quantity. It is defined as:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time elapsed

Acceleration is measured in meters per second squared (m/s^2). It must be noticed that acceleration is a vector, so it also has a direction. In particular:

- when acceleration is negative, it means that the object is slowing down, so acceleration is in opposite direction to the velocity

- when acceleration is positive, it means that the object is speeding up, so acceleration is in the same direction as the velocity

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3 years ago