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3 years ago

Which wave has a higher frequency than microwaves but lower frequency than UV waves?

Physics

1 answer:

Effectus [21]3 years ago

3 0

Answer:

Infrared

Visible light

Explanation:

Electromagnetic waves are waves consisting of oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion of the wave.

Electromagnetic waves travel in a vacuum at the speed of light, which is a constant value:

$c=3.00\cdot 10^8 m/s$

Also, electromagnetic waves are classified into 7 different types, depending on their wavelength and frequency. From highest to lowest frequency, we have:

gamma-rays $10^{20}-10^{24} Hz$

x-rays $10^{17}-10^{20}Hz$

ultraviolet $10^{15}-10^{17} Hz$

visible $[4-7.5]\cdot 10^{14} Hz$

infrared $10^{14} Hz$

microwaves $10^{11}-10^{13} Hz$

radio waves

Therefore, from the table we see that both infrared and visible light have higher frequency than microwaves, but lower frequency than UV (ultraviolet).

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Determine the amount of time for polonium-210 to decay to one fourth its original quantity. The half-life of polonium-210 is 138

Tju [1.3M]

Answer:

276 days

Explanation:

1/4 th of the original means 2 half lives

1 half life = 138 days

So,

2 half lives = 276 days

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3 years ago

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3 years ago

For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103

Alona [7]

Answer:

a) P = 44850 N

b) $\delta l =0.254\ mm$

Explanation:

Given:

Cross-section area of the specimen, A = 130 mm² = 0.00013 m²

stress, σ = 345 MPa = 345 × 10⁶ Pa

Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa

Initial length, L = 76 mm = 0.076 m

a) The stress is given as:

$\sigma=\frac{\textup{Load}}{\textup{Area}}$

on substituting the values, we get

$345\times10^6=\frac{\textup{Load}}{0.00013}$

Load, P = 44850 N

Hence the maximum load that can be applied is 44850 N = 44.85 KN

b)The deformation ( $\delta l$ ) due to an axial load is given as:

$\delta l =\frac{PL}{AE}$

on substituting the values, we get

$\delta l =\frac{44850\times0.076}{0.00013\times103\times 10^9}$

$\delta l =0.254\ mm$

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3 years ago

A pickup truck is carrying a 10.0 kg toolbox, but the tailgate of the truck is missing, so the box can slide out if it starts mo

scoray [572]

Answer: The shortest time the truck could take is 4.34 secs

Explanation: Please see the attachments below

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3 years ago

Calculate how much work is done when a force of 1600 N pulls a car through a distance of 40 m in the direction of the force.

Ulleksa [173]

Answer: work done = 64,000J or 64KJ.

Explanation:

Work is said to be done if a force applied makes the object to move through a distance in the direction of the applied force. Work is the product of force and perpendicular distance in the direction of the applied force. It is represented by W, measured in Joules(J) and a scalar quantity.

Thus, work done(J)= force(N) × distance(m)

From the the question,

applied force= 1600N

distance=40m

Therefore, work= 1600 × 40

W= 64,000J

But, 1 kilojoules = 1000J

Therefore, 64,000J = 64,000 ÷ 1000

= 64KJ

Work done is 64KJ

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3 years ago