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3 years ago

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What is the process in which the work of scientists is evaluated by other researchers? a. Scientific critique c. scientific inqu

iry b. Peer review d. Inquisition

2 answers:

Natali5045456 [20]3 years ago

7 0

The answer to your question would be peer review

prisoha [69]3 years ago

4 0

Answer:

Explanation:

The process in which the work of scientists is evaluated by other researchers is called as Peer review.

Peer review is one of the gold standards of science. It is a method in which scientists, “peers” value the standard or the quality of fellow scientists’ work. By doing so, they aim to make sure the work is rigorous, coherent, relevant and uses past analysis and adds to what we have already known. All in all peer review improves the quality of work.

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Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity of +4.0 m/s a

DedPeter [7]

Answer:

As collision is elastic,thus we can use conservation of momentum equation

mA=0.2 kg

(vB)1=0 m/s.......................as it is on rest before collision

(vA)1=4 m/s

(vA)2=-1 m/s

(vB)2=2 m/s

using equation

(mA*vA+mB*vB)1= (mA*vA+mB*vB)2

Where 1 and 2 represents before and after collision

(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)

0.8=-0.2+(2mB)

mass of object B=mB=0.3 Kg

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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:

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

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3 years ago

When a wave is acted upon by an external damping force, what happens to the energy of the wave

Rasek [7]

The energy of the wave decreases gradually

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3 years ago

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Please help me with this

aev [14]

Answer:

the answer is UV Radiation

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3 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

Sati [7]

(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

with horizontal and vertical components, respectively,

$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

The horizontal component of the ball's velocity is constant throughout its trajectory, so $v_{x,i}=v_{x,f}$ , and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time <em>t</em> is

$v_{y,f}=v_{y,i}-gt$

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

So, the initial velocity vector is

$\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

and direction <em>θ</em> such that

$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

$y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}$

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3 years ago