Answer:
the wind carries abrasive materials
Explanation:
such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first
A <u>scanner</u> is a type of communications equipment that functions as a radio receiver and searches across several frequencies.
A scanner is a kind of a radio receiver that has the ability to receive multiple signals.
There are three modes which a scanner uses for acting as a radio receiver. The scan mode of the radio receiver constantly changes frequencies that helps in transmissions. There is also a manual scan mode that allows the users to search for their interested frequencies. The search mode allows the users to search through two sets of frequencies.
A scanner is a type of communication equipment that is easy to use with various features such as the volume, numeric keypad, trunk tracking etc.
To learn more about scanners, click here:
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Answer:
200000 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) of roller coaster = 1000 Kg
Velocity (v) of roller coaster = 20 m/s
Kinetic energy (KE) =?
Kinetic energy is simply defined as the energy possess by an object in motion. Mathematically, it can be expressed as:
KE = ½mv²
Where
KE => is the kinetic energy.
m =>is the mass of the object
V => it the velocity of the object.
With the above formula, we can obtain the kinetic energy of the roller coaster as follow:
Mass (m) of roller coaster = 1000 Kg
Velocity (v) of roller coaster = 20 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 1000 × 20²
KE = 500 × 400
KE = 200000 J
Therefore, the kinetic energy of the roller coaster is 200000 J.
Answer:
v = 29.4 m / s
Explanation:
For this exercise we can use the conservation of mechanical energy
Lowest starting point.
Em₀ = K = ½ m v²
final point. Higher
= U = m g h
Let's use trigonometry to lock her up
cos 60 = y / L
y = L cos 60
Height is the initial length minus the length at the maximum angle
h = L - L cos 60
h = L (1- cos 60)
energy is conserved
Em₀ = Em_{f}
½ m v² = mgL (1 - cos 60)
v = 2g L (1- cos 60)
let's calculate
v² = 2 9.8 3.0 (1- cos 60)
v = 29.4 m / s
