Answer:
An <u>applied force</u> is a force that is applied to an object by a person or another object. If a person is pushing a desk across the room, then there is an applied force acting upon the object. The applied force is the force exerted on the desk by the person.
A <u>friction force</u> is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing intermolecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an object can be calculated using the formula below:
= µ •
Answer:
λ = 3.2 x 10⁻⁷ m = 320 nm
Explanation:
The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:
v = fλ
where,
v = c = speed of the electromagnetic waves (UV rays) = speed of light
c = 3 x 10⁸ m/s
f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz
λ = wavelength of the electromagnetic waves (UV rays) = ?
Therefore, substituting the values in the relation, we get:
3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)
λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)
<u>λ = 3.2 x 10⁻⁷ m = 320 nm</u>
So, the radiation of <u>320 nm</u> wavelength is absorbed by Ozone.
Downward movement under the force of gravity only.
Answer:
t = 5.05 s
Explanation:
This is a kinetic problem.
a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m
b) in this system the equations of motion are
y = v₀ t + ½ g t²
where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth
e) y = 0 + ½ g t²
t = √ (2y / g)
t = √(2 125 / 9.8)
t = 5.05 s