S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)
S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m
Distance double 720m*2=1440m
V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
infiltration and evaporation
Answer:
854.39 N
Explanation:
The formula for the fundamental frequency of a stretched string is given as,
f = 1/2L√(T/m)..................... Equation 1
Where f = fundamental frequency, L = Length of the wire, T = Tension, m = mass per unit length.
Given: f = 261.6 Hz, L = 0.6 m, m = (5.2×10⁻³/0.6) = 8.67×10⁻³ kg/m.
Substitute into equation 1
261.6 = 1/0.6√(T/8.67×10⁻³)
Making T the subject of the equation,
T = (261.6×0.6×2)²(8.67×10⁻³)
T =854.39 N
Hence the tension of the wire is 854.39 N.
<u>2352 Joules</u> is the “Gravitational potential energy” that a skater has.
<u>Explanation</u>:
"Gravitational potential energy" of an object is defined as the energy stored in the object which is at a height. Everybody at a height has some energy that is calculated by using the potential energy formula. Mathematically, "Gravitational potential energy" (GPE) is mass (m) times acceleration due to gravity (g) times height (h) of an object. Therefore, GPE = m × g × h.
From the given question,
Mass of the skater is 60 kg
The height at which skater is located is 4 m
“g” is acceleration due to gravity 9.8
Substitute the given values in the GPE = m × g × h
Gravitational potential energy = 60 × 9.8 × 4
Gravitational potential energy = 2352 J
.
Answer:
The wire meet the ground at an angle of 56.4 degrees
Explanation:
It is given that,
To support a tree damaged in a storm, a 12-foot wire is secured from the ground to the tree at a point 10 feet off the ground.
The hypotenuse is, H = 12 foot
The perpendicular distance is, P = 10 feet
The angle between the tree and the ground is 90 degrees
Using Pythagoras theorem as :
So, the wire meet the ground at an angle of 56.4 degrees. Hence, the correct option is (d).