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2 years ago

Short-range forecasts tends to ________ longer-range forecasts.

1 answer:

5 0

Short-range forecasts are more accurate than longer range ones. Short term forecasts may use mathematical techniques such as moving averages and, exponential smoothing. Longer term forecasts not only use different methodologies, such as qualitative vs. quantitative, they also tend to consider different issues.

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A) The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion

suter [353]

A) The mass of the continent is $2.5\cdot 10^{21} kg$

B) The kinetic energy is 2016 J

C) The speed of the jogger should be 7.1 m/s

Explanation:

The mass of the continent can be calculated as

$m = \rho V$

where

$\rho = 2800 kg/m^3$ is its density

V is its volume

We have to calculate its volume. We know that the continent is represented as a slab of side 5900 km (so its surface is 5900 x 5900, assuming it is a square) and depth of 26 km, so its volume is:

$V=(5900 km)^2 (26 km)=9.05\cdot 10^8 km^3 =9.05 \cdot 10^8 \cdot (10^9 m^3/k^3)=9.05\cdot 10^7 m^3$

So, the mass of the continent is

$m=\rho V = (2800)(9.05\cdot 10^{17})=2.5\cdot 10^{21} kg$

The kinetic energy of a body is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the continent, we have:

$m=2.5\cdot 10^{21} kg$ is the mass

$v=4 cm/year$ is the speed

We have to convert the speed into SI units. we have:

1 cm = 0.01 m

$1 year = (365)(24)(60)(60) s = 3.15\cdot 10^7 s$

So, the speed is

$v=4 cm/year = 0.04 m/year \cdot \frac{1}{3.15\cdot 10^7}=1.27\cdot 10^{-9} m/s$

Therefore, the kinetic energy is

$K=\frac{1}{2}(2.5\cdot 10^{21} kg)(1.27\cdot 10^{-9} m/s)^2=2016 J$

Again, the kinetic energy of an object is

$K=\frac{1}{2}mv^2$

For the jogger in this problem, his mass is

m = 80 kg

And we want its kinetic energy to be equal to that of the continent, so

K = 2016 J

Re-arranging the equation for v, we find what speed the jogger needs to have this kinetic energy:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2016)}{80}}=7.1 m/s$

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

8 0

2 years ago

A glider of mass m1 on a frictionless horizontal track is connected to an object of mass m2 by a massless string. The glider acc

inysia [295]

Answer:

m2g -> T2 -> T1

Explanation:

m2g -> T2 -> T1

5 0

2 years ago

Force X has a magnitude of 1260 pounds, and Force Y has a magnitude of 1530 pounds. They act on a single point at an angle of 4

weeeeeb [17]

Answer:

Fe= 2579.68 P

α= 24.8°

Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

FX = 1260 P , horizontal (-x)

FY = 1530 P , forming 45° with positive x axis

x-y components FY

FYx= - 1530*cos(45)° = - 1081.87 P

FYy= - 1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

Fny=FYy

Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny= -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

$F_{n} = \sqrt{(F_{nx})^{2} +(F_{ny})^{2} }$

$F_{e} =\sqrt{(2341.87)^{2}+(1081.87)^{2} }$

Fe= 2579.68 P

Calculation of the direction of equilibrant force (α)

$\alpha =tan^{-1} (\frac{F_{ny} }{F_{nx} } )$

$\alpha =tan^{-1} (\frac{1081.87 }{2341.87} )$

α= 24.8°

Look at the attached graphic

6 0

2 years ago

While you are studying for an upcoming physics exam, a lightning storm is brewing outside your window. Suddenly, you see a tree

vovikov84 [41]

Blahbsnansjsjsjsisisisisiskskssk

7 0

2 years ago

The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan

PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire. If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

Explanation:

Magnetic field around a long current carrying wire is given by

$B=\frac{\mu _o I}{2\pi r}$

where B= magnetic field

$\mu _o=$ permeability of free space

I= current in the long wire and

r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is directly proportional to the current and inversely proportional to the distance from the wire.

Now if I'=3I and r'=2r then magnetic field B' is given by

$B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B$

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by one and half (1.5) times.

7 0

2 years ago

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