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3 years ago

Radiation is an example of a transverse wave t or f

Physics

1 answer:

s2008m [1.1K]3 years ago

5 0

That's true for electromagnetic radiation.

(Not for some other kinds though, like alpha radiation and beta radiation. Those are particles, not waves at all.)

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A thunderclap sends a sound wave through the air and the ocean below. The

marysya [2.9K]

Answer:

C. 14.93 m

Explanation:

The given frequency of the wave, f = 100 Hz

The given equation for the wave speed, <em>v</em>, is presented as follows;

v = f × λ

The speed of sound in water, v = 1,493 m/s

Therefore, we get;

The wavelength, λ = v/f

∴ λ = 1,493 m/s/(100 Hz) = 14.93 m

The wavelength, λ = 14.93 m.

8 0

3 years ago

What is the definition of the half-life of a radioactive isotope?

Illusion [34]

Answer:Half-life is the amount of time it takes for the initial mass of the isotope to decompose, by half, into other lighter atoms.

Explanation:Different radioactive isotopes have different half-lives. For example, the element technetium-99m has a half life of 6 hours. This means that is 100 kg of the element is left to decay, in 6 hours, 50kg of the mass will have changed into other elements/atoms. The half-life of uranium-238 is 4.5 billion years while that of polonium-216 is only 0.145 seconds.

5 0

2 years ago

Read 2 more answers

A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0

3 years ago

A 7950-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s . A 2950-kg load, initially

Alchen [17]

The new speed of car is 10.9 m/s

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According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

$m_1u_1 + m_2u_2 = (m_1 + m_2) v$