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3 years ago

12

in December which will have a longer day length a country im notheren hemisphere or southern hemisphere

1 answer:

astra-53 [7]3 years ago

3 0

Answer:

Southern Hemisphere

Explanation:

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What kind of cord is this no links please

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Explanation:

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2 years ago

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What is the length of the X component of the vector plotted below

dlinn [17]

Answer:

5

Explanation:

If you straighten out the line, it touches the 5, which makes the length 5

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2 years ago

the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t

givi [52]

Explanation:

Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.

→ The tangential component of acceleration is rate of increase in the speed of plane so,

$a_{t} = v = 0.8 m/s^{2}$

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).

dy/dx = d(0.4x²)/dx

= 0.8x

Take the derivative again,

d²y/dx² = d(0.8x)/dx

= 0.8

at x= 5 Km

dy/dx = 0.8(5)

= 4

$p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }$

now insert the values,

$p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km$

→ Now the normal component of acceleration is given by

$a_{n} = \frac{v^{2} }{p}$

= (200)²/(87.6×10³)

aₙ = 0.457 m/s²

→ Now the total acceleration is,

$a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}$

$a = [(0.8)^{2} + (0.457)^{2}]^{0.5}$

a = 0.921 m/s²

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2 years ago

Define vector quantity giving examples

Mama L [17]

Answer:

Quantity with direction and magnitude

Displacement

Explanation:

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2 years ago

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

muminat

Answer:

$2.1\cdot 10^{21}$ electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a charged sphere, so we have:

$E_b=3.00\cdot 10^6 N/C$ is the maximum electric field strength tolerated by the air before breakdown occurs

$r=1.00 km = 1000 m$ is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C$

Assuming that the cloud is negatively charged, then

$Q=-333.3 C$

And since the charge of one electron is

$e=-1.6\cdot 10^{-19}C$

The number of excess electrons on the cloud is

$N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}$

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2 years ago