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3 years ago

One property that makes electromagnetic waves differ from other types of waves is that they can

Physics

2 answers:

AysviL [449]3 years ago

8 0

The correct answer is B.travel in a vacuum.

Phoenix [80]3 years ago

4 0

That they travel in a vacuum. All other waves require a medium in which they wave.

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How fast can a 4000 kg truck travel around a 70 m radius turn without skidding if its tires share a 0.6 friction coefficient wit

aleksley [76]

Answer:

Velocity of truck will be 20.287 m /sec

Explanation:

We have given mass of the truck m = 4000 kg

Radius of the turn r = 70 m

Coefficient of friction $\mu =0.6$

Centripetal force is given $F=\frac{mv^2}{r}$

And frictional force is equal to $F_{frictional}=\mu mg$

For body to be move these two forces must be equal

So $\frac{mv^2}{r}=\mu mg$

$v=\sqrt{\mu rg}=\sqrt{0.6\times 70\times 9.8}=20.287m/sec$

7 0

3 years ago

A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force

salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

$KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J$

Energy lost due to friction

$W=Fd=0.031\times 0.158=0.0048J$

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

$\frac{1}{2}mv^2=0.0052$

$\frac{1}{2}\times 0.00524\times v^2=0.0052$

v = 1.40 m/sec

7 0

3 years ago

A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th

s344n2d4d5 [400]

Answer:

15 cm

Explanation:

$h_{o}$ = Diameter of the coin = 15 mm

$h_{i}$ = Diameter of the image of coin = 5 mm

$d_{o}$ = distance of the coin from mirror = 15 cm

$d_{i}$ = distance of the image of coin from mirror = ?

Using the equation

$\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}$

$\frac{d_{i}}{15} = \frac{- (5)}{15}$

$d_{i}$ = - 5 cm

$R$ = radius of curvature

Using the mirror equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}$

$\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}$

$R$ = - 15 cm

6 0

3 years ago

Classify the type of plate boundary where the appalachian mountains formed . how have they changed since their formation

disa [49]

<span>The Appalachian Mountains were formed when colliding tectonic plates folded and upthrust, mainly during the Permian Period and again in the Cretaceous Period. The folds and thrusts were then eroded and carved by wind, streams and glaciers. These erosive processes are ongoing, and the topography of the Appalachian Mountains continue to change. They have changed with the miles of land that are cleared of all vegetation and topsoil. In the 1970's coal miners literally blow away the top of a mountain to get to the coal underneath.</span>

6 0

3 years ago

A planet is discovered orbiting around a star in the galaxy Andromeda at four times the distance from the star as Earth is from

Dmitriy789 [7]

Answer:

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

Explanation:

The orbital period of a planet around a star can be expressed mathematically as;

T = 2π√(r^3)/(Gm)

Where;

r = radius of orbit

G = gravitational constant

m = mass of the star

Given;

Let R represent radius of earth orbit and r the radius of planet orbit,

Let M represent the mass of sun and m the mass of the star.

r = 4R

m = 16M

For earth;

Te = 2π√(R^3)/(GM)

For planet;

Tp = 2π√(r^3)/(Gm)

Substituting the given values;

Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)

Tp = 2π√(4R^3)/(GM)

Tp = 2 × 2π√(R^3)/(GM)

So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

5 0

3 years ago

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