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2 years ago

Consider the following:

Physics

1 answer:

pychu [463]2 years ago

5 0

Answer:

They have different wavelengths.

They have different frequencies.

They propagate at different speeds through non-vacuum media depending on both their frequency and the material in which they travel.

Explanation:

The complete question is

Consider the following:

a) radio waves emitted by a weather radar system to detect raindrops and ice crystals in the atmosphere to study weather patterns;

b) microwaves used in communication satellite transmissions;

c) infrared waves that are perceived as heat when you turn on a burner on an electric stove;

d) the multicolor light in a rainbow;

e) the ultraviolet solar radiation that reaches the surface of the earth and causes unprotected skin to burn; and

f) X rays used in medicine for diagnostic imaging.

Which of the following statements correctly describe the various forms of EM radiation listed above?

check all that apply to the above

They have different wavelengths.

They have different frequencies.

They propagate at different speeds through a vacuum depending on their frequency.

They propagate at different speeds through non-vacuum media depending on both their frequency and the material in which they travel.

They require different media to propagate.

All the above phenomena are due the electromagnetic wave spectrum. Electromagnetic waves travel at a constant speed of 3 x 10^8 m/s in a vacuum. Within the spectrum, the different types of electromagnetic waves exists in different band range of frequencies and wavelengths unique to each of the waves, and the energy they carry. When these waves enter a non-vacuum medium, their speed change, depending on the nature of the material of the medium, and the frequency or the wavelength of the incoming wave.

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An 80-g particle moving with an initial speed of 50 m/s in the positive x direction strikes and sticks to a 60-g particle moving

liubo4ka [24]

The collision is a form of inelastic collision because the it forms a single mass after is collides. So it can be solve by momentum balance

( 0.08 kg * 50 m/s ) + ( 0.06 kg * 50 m/s) = ( 0.08 + 0.06 kg ) v

V = 50 m/s

So the kinetic energy lost is

KE = 0.5 (50 m/s)^2) *( 0.14 – 0.08kg )

KE = 75 J

8 0

2 years ago

The force F required to compress a spring a distance x is given by F 2 F0 5 kx where k is the spring constant and F0 is the prel

IrinaVladis [17]

Answer:

a)W=8.333lbf.ft

b)W=0.0107 Btu.

Explanation:

<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

Preload = F₀=0 lbf

Spring constant k= 200 lbf/in

Initial length of spring x₁=0

Final length of spring x₂= 1 in

At any point, the force during deflection of a spring is given by;

F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

$W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2} )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\$

Change to lbf.ft by dividing the value by 12 because 1ft=12 in

100/12 = 8.333 lbf.ft

work required to compress the spring, W=8.333lbf.ft

The work required to compress the spring in Btu will be;

1 Btu= 778 lbf.ft

?= 8.333 lbf.ft----------------cross multiply

(8.333*1)/ 778 =0.0107 Btu.

6 0

2 years ago

A ship maneuvers to within 2.50 x 103 m of anisland's 1.80 x 103 m high mountain peak and fires aprojectile at an enemy ship 6.1

const2013 [10]

Answer:

Distance between peak height (vertically) of projectile and mountain height = (2975.2 - 1800) = 1175.2 m

Distance between where the projectile lands and ship B = (3188.8 - 3110) = 8.8 m

Explanation:

Given the velocity and angle of shot of the projectile, one can calculate the range and maximum height attained by the projectile.

H = (v₀² Sin²θ)/2g

v₀ = initial velocity of projectile = 2.50 × 10² m/s = 250 m/s

θ = 75°, g = 9.8 m/s²

H = 250² (Sin² 75)/(2 × 9.8) = 2975.2 m

Range of projectile

R = v₀² (sin2θ)/g

R = 250² (sin2×75)/9.8

R = 250² (sin 150)/9.8 = 3188.8 m

Height of mountain = 1.80 × 10³ = 1800 m

Maximum height of projectile = 2975.2 m

Distance between peak height (vertically) of projectile and mountain height = 2975.2 - 1800 = 1175.2 m

Distance of ship B from ship A = 2.5 × 10³ + 6.1 × 10² = 2500 + 610 = 3110 m

Range of projectile = 3188.8 m

Distance between where the projectile lands and ship B = 3188.8 - 3110 = 8.8 m

8 0

2 years ago

A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays

IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

$p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s$

Final momentum

$p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s$

4 0

2 years ago

Which statements describe the applications of nuclear medicine scans? Check all that apply.

coldgirl [10]

Answer:

- asses disease progression and tissue function

- utilize a biologically active molecule

- utilize a radionuclide tracer

Explanation:

4 0

2 years ago

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