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3 years ago

What do scientist do to receive suggestions an criticism of their research from other scientists?

Physics

2 answers:

Nady [450]3 years ago

8 0

Answer:

they publish

Explanation:

barxatty [35]3 years ago

4 0

They PUBLISH !

They provide complete written, detailed, technical descriptions of their thoughts, hypotheses, experiments, data, and conclusions, in publications that are read by other scientists around the world.

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which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20

MA_775_DIABLO [31]

<span>Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
Plug in the values in equation (A):
20 = 10 * a
Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
Mass = m = 18 kg
Plug in the values in equation (A):

20 = 18 * a
Acceleration = a = 1.11 m/s^2

2 > 1.11; therefore, 10 kg object has the higher acceleration compared to the acceleration of the 18 kg object.</span>

7 0

3 years ago

Read 2 more answers

Does anyone know the answer

scoundrel [369]

No put false I am right

7 0

2 years ago

The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum

Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.

5 0

2 years ago

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10r, where a and fare in m/s^2

xz_007 [3.2K]

Answer:

Mistake in question

The correct question

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10t , where a and t are in m/s² and seconds, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.02v², where v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to 10 m/s, (b) the total distance the car travels during this time.

Explanation:

Given the function

a = 50 —10t

The car started from rest u = 0

And it accelerates to a speed of 125m/s

Then, let find the time in this stage

Acceleration can be modeled by

a = dv/dt

Then, dv/dt = 50—10t

Using variable separation to solve the differentiation equation

dv = (50—10t)dt

Integrating both sides

∫ dv = ∫ (50—10t)dt

Note, v ranges from 0 to 125seconds, so we want to know the time when it accelerate to 125m/s. So t ranges from 0 to t'

∫ dv = ∫ (50—10t)dt

v = 50t —10t²/2. Equation 1

[v] 0<v<125 = 50t —10t²/2 0<t<t'

125—0 = 50t — 5t² 0<t<t'

125 = 50t' — 5t'²

Divide through by 5

25 = 10t' — t'²

t'² —10t' + 25 = 0

Solving the quadratic equation

t'² —5t' —5t' + 25 = 0

t'(t' —5) —5(t' + 5) = 0

(t' —5)(t' —5) = 0

Then, (t' —5) = 0 twice

Then, t' = 5 seconds twice

So, the car spent 5 seconds to get to 125m/s.

The second stage when the parachute was deployed

We want to the time parachute reduce the speed from 125m/s to 10m/s,

So the range of the velocity is 125m/s to 10m/s. And time ranges from 0 to t''

The function of deceleration is give as

a = - 0.02v²

We know that, a = dv/dt

Then, dv/dt = - 0.02v²

Using variable separation

(1/0.02v²) dv = - dt

(50/v²) dv = - dt

50v^-2 dv = - dt

Integrate Both sides

∫ 50v^-2 dv = -∫dt

(50v^-2+1) / (-2+1)= -t

50v^-1 / -1 = -t

- 50v^-1 = -t

- 50/v = - t

Divide both sides by -1

50/v = t. Equation 2

Then, v ranges from 125 to 10 and t ranges from 0 to t''

[ 50/10 - 50/125 ] = t''

5 - 0.4 = t''

t'' = 4.6 seconds

Then, the time taken to decelerate from 125s to 10s is 4.6 seconds.

So the total time is

t = t' + t''

t = 5 + 4.6

t = 9.6 seconds

b. Total distanctraveleded.

First case again,

We want to find the distance travelled from t=0 to t = 5seconds

a = 50—10t

We already got v, check equation 1

v = 50t —10t²/2 + C

v = 50t — 5t² + C

We add a constant because it is not a definite integral

Now, at t= 0 v=0

So, 0 = 0 - 0 + C

Then, C=0

So, v = 50t — 5t²

Also, we know that v=dx/dt

Therefore, dx/dt = 50t — 5t²

Using variable separation

dx = (50t —5t²)dt

Integrate both sides.

∫dx = ∫(50t —5t²)dt

x = 50t²/2 — 5 t³/3 from t=0 to t=5

x' = [25t² — 5t³/3 ]. 0<t<5

x' = 25×5² — 5×5³/3 —0

x' = 625 — 208.333

x' = 416.667m

Stage 2

The distance moved from

t=0 to t =4.6seconds

a = -0.002v²

We already derived v(t) from the function above, check equation 2

50/v = t + C.

When, t = 0 v = 125

50/125 = 0 + C

0.4 = C

Then, the function becomes

50/v = t + 0.4

50v^-1 = t + 0.4

Now, v= dx/dt

50(dx/dt)^-1 = t +0.4

50dt/dx = t + 0.4

Using variable separation

50/(t+0.4) dt = dx

Integrate both sides

∫50/(t+0.4) dt = ∫ dx

50 In(t+0.4) = x

t ranges from 0 to 4.6seconds

50In(4.6+0.4)—50In(4.6-0.4) = x''

x'' = 50In(5) —50In(4.2)

x'' = 8.72m

Then, total distance is

x = x' + x''

x = 416.67+8.72

x = 425.39m

The total distance travelled in both cases is 425.39m

5 0

2 years ago

Read 2 more answers

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

$\theta=5.71^{o}$

8 0

2 years ago