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LekaFEV [45]
3 years ago
6

Why are some scientists skeptical of the scientists results and conclusions

Physics
1 answer:
babunello [35]3 years ago
6 0

Earth's surface temperature has warmed 1.5 degrees Celsius over the past 250 years and "humans are almost entirely the cause," concludes a scientific study designed to address skeptical concerns that the causes of climate change are even induced. for the man. Professor Richard Muller, a climate change physicist and founder of the Berkeley Earth Surface Temperature Project, said he was surprised by the findings. "We did not expect this result, but as scientists it is our duty to let the evidence change our minds." He added that he now considers himself a "converted skeptic" and his views have been subjected to a "total turnaround" in a short time. "Our results show that the average land surface temperature has increased by 1.5 degrees Celsius over the last 250 years, including an increase of 0.9 degrees Celsius over the last 50 years. In addition, it seems likely that essentially all of this increase results from human emissions of greenhouse gases, ”Muller wrote in an article in the New York Times. The University of California, Berkeley-based team of scientists has assembled and consolidated a set of 14.4 million surface temperature observations, collected from 44,455 sites worldwide, dating back to 1753. Previous datasets created by NASA and the National Oceanic and Atmospheric Administration (NOAA) in the US, and by the Met Office and the Department of Climate Research at the University of East Anglia in the United Kingdom only arrived until the mid-19th century and used only a fifth of the number. of weather station records

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An object, initially at rest moves 250m in 17s. What is it's acceleration?
Gekata [30.6K]
With the values you've given, only velocity can be found.
Acceleration is rate of change of velocity

d= 250s
t= 17s

a= d/t
 =\frac{250}{17}
 = 4.7 \frac{m}{s}
8 0
2 years ago
The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 50 km/
velikii [3]

To solve this problem, you must figure out (in vector form) both the wind vector and plane vector

w⃗ = wind vector

P⃗ = plane vector

 To get the  true course of the plane, you need to add the plane and wind vectors, the formula would be

w⃗ +P⃗ ,

 which will result to the ground speed.

ground speed=||w⃗ +P⃗ ||

 

Using the planar representation of your situation, this will help you understand the equation, use this to make the equation more understandable.

w⃗ =AB¯¯¯¯¯¯¯¯,   P⃗ =AC¯¯¯¯¯¯¯¯

 

the smaller circle is of radius 50 (similar to the wind speed) and the larger circle is of radius 200 (similar to the plane vector.  To get the coordinates of these two vectors,  use polar coordinates.

Let East be 0 degrees, so since the wind vector is on the circle of radius 50, we have:

w⃗ =⟨50cos(135),50sin(135)⟩=⟨−252√, 252√⟩.

P⃗ =⟨200cos(60),200sin(60)⟩=⟨100,\1003√⟩.

w⃗ +P⃗ =⟨100−252√ , 1003√+252√⟩

||w⃗ +P⃗ ||=(100−252√)2+(1003√+252√)2

√≈218.349218.

 

<span> </span>

8 0
2 years ago
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n
satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0
3 years ago
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.
Veronika [31]

Answer:

A) ω = 6v/19L

B) K2/K1 = 3/19

Explanation:

Mr = Mass of rod

Mb = Mass of bullet = Mr/4

Ir = (1/3)(Mr)L²

Ib = MbRb²

Radius of rotation of bullet Rb = L/2

A) From conservation of angular momentum,

L1 = L2

(Mb)v(L/2) = (Ir+ Ib)ω2

Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.

(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2

(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2

Divide each term by Mr;

vL/8 = (L²/3 + L²/16)ω2

vL/8 = (19L²/48)ω2

Divide both sides by L to obtain;

v/8 = (19L/48)ω2

Thus;

ω2 = 48v/(19x8L) = 6v/19L

B) K1 = K1b + K1r

K1 = (1/2)(Mb)v² + Ir(w1²)

= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)

= (1/8)(Mr)v²

K2 = (1/2)(Isys)(ω2²)

I(sys) is (Ir+ Ib). This gives us;

Isys = (19L²Mr/48)

K2 =(1/2)(19L²Mr/48)(6v/19L)²

= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152

Thus, the ratio, K2/K1 =

[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19

3 0
3 years ago
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the
igomit [66]

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

y = \frac{\lambda L}{d}\\\\d  =\frac{\lambda L}{y}\\\\

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>

7 0
3 years ago
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