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3 years ago

Which part of the electromagnetic spectrum has a lower frequency than visible light?

Physics

1 answer:

Anastasy [175]3 years ago

5 0

Answer:

radio waves

Explanation:

radio waves have lower frequency than visible light

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If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate

pogonyaev

Let h = distance (m) to the water surface.

Initial velocity, u = 0 (because the stone was dropped).
Use the formula
h = ut + (1/2)gt^2
where g = 9.8 m/s^2 (acc. due to graity)
t = time (s)

h = (1/2)*(9.8)*(3^2) = 44.1 m

8 0

3 years ago

In certain cases, using both the momentum principle and energy principle to analyze a system is useful, as they each can reveal

kramer

Explanation:

The gravitational force equation is the following:

$F_G = G * \frac{m_1 m_2}{r^2} \\$

Where:

G = Gravitational constant = $6.67408 * 10^{-11} m^3 kg^{-1} s^{-2}$

m1 & m2 = the mass of two related objects

r = distance between the two related objects

The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.

8 0

3 years ago

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$