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3 years ago

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Physics

2 answers:

adelina 88 [10]3 years ago

4 0

Answer:

Explanation:,

Aneli [31]3 years ago

3 0

Answer:Climatic processes affect the dynamics of Earth's ice sheets and glaciers, and along ... by abrupt events and by continuous reshaping of Earth's surface from surface ... Forecasting natural disasters, including the timing and size of earthquakes, the . Last, human activity has a profound impact on water resources, landscape

Explanation:

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Through both refraction and diffraction, the atmosphere alters the apparent speed and, to a lesser extent, the direction of the

musickatia [10]

Answer:

The Ionospheric Effect

Explanation:

One of the largest errors in GPS positioning is attributable to the atmosphere. The long, relatively unhindered travel of the GPS signal through the virtual vacuum of space changes as it passes through the earth’s atmosphere. Through both refraction and diffraction, the atmosphere alters the apparent speed and, to a lesser extent, the direction of the signal. This causes an apparent delay in the signal's transit from the satellite to the receiver.

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3 years ago

. Maria walked 1.5 miles south to her house in 0.5 hours. What is her speed in miles per hour?

Simora [160]

1) 3 miles/Hour

The speed is defined as the distance covered divided by the time taken:

$v=\frac{d}{t}$

where

d = 1.5 mi is the distance

t = 0.5 h is the time taken

Substituting,

$v=\frac{1.5}{0.5}=3 mi/h$

2) 1.34 m/s south

Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:

$d=1.5 mi \cdot 1609 m/mi = 2414 m$ is the displacement in meters

$t=0.5 h \cdot 3600 s/h =1800 s$ is the time taken in seconds

Substituting,

$v=\frac{2414 m}{1800 s}=1.34 m/s$

And the direction of the velocity is the same as the displacement, so it is south.

6 0

4 years ago

A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and ou

zepelin [54]

Explanation:

Given that,

Radius R= 2.00

Charge = 6.88 μC

Inner radius = 4.00 cm

Outer radius = 5.00 cm

Charge = -2.96 μC

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{kq}{r^2}$

(a). For, r = 1.00 cm

Here, r<R

So, E = 0

The electric field does not exist inside the sphere.

(b). For, r = 3.00 cm

Here, r >R

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}$

$E=6.88\times10^{7}\ N/C$

The electric field outside the solid conducting sphere and the direction is towards sphere.

(c). For, r = 4.50 cm

Here, r lies between R₁ and R₂.

So, E = 0

The electric field does not exist inside the conducting material

(d). For, r = 7.00 cm

The electric field is

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}$

$E=5.43\times10^{6}\ N/C$

The electric field outside the solid conducting sphere and direction is away of solid sphere.

Hence, This is the required solution.

6 0

3 years ago

Two long straight wires are parallel and carry current in the same direction. The currents are 8.0 and 12 A and the wires are se

Bess [88]

A magnetic field is a mathematical description of the magnetic influence of electric currents and magnetic materials. The magnetic field at any point is specified by two values, the direction and the magnitude; such that it is a vector field. Mathematically it is described as,

$B = \frac{\mu_0 I}{2\pi d}$