Answer:
(a) The net force is 80.394 N
The acceleration of the crate is 0.804 m/s²
(b) the final velocity of the crate is 5.02 m/s
Explanation:
Given;
mass of the crate, m = 100 kg
applied force, F = 250 N
angle of inclination, θ = 45°
coefficient of friction, μ = 0.12
Applied force in y-direction, 
Applied force in x-direction, 
The normal force is calculated as;
N + Fy -W = 0
N = W - Fy
N = (100 x 9.8) - 176.78
N = 980 - 176.78 = 803.22 N
The frictional force is given by;
Fk = μN
Fk = 0.12 x 803.22
Fk = 96.386 N
(a) The net force is given by;
Apply Newton's second law of motion;
F = ma
(b) the velocity of the crate after 5.0 s
Answer:
Water is more dense than air. When water goes through a denser thing, the light is "bent" more towards the "normal" which is a straight, vertical line.
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Answer:
Explanation:
To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .
I₁ = 1/2 mr²
= .5 x 175 x 2.13²
= 396.97 kgm²
I₂ = m r²
= 55.4 x 2.13²
= 251.34 mgm²
ω₁ = .651 rev /s
= .651 x 2π rad /s
ω₂ = tangential velocity of man / radius of disc
= 3.51 / 2.13
= 1.65 rad/s
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω
ω = 3.14 rad /s
kinetic energy = 1/2 I ω²
= 3196 J
The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.
Em = K + U
Let's write the energy in two points.
Starting point. Highest part of the oscillation
Em₀ = U = m g h
Final point. Lower part of the movement
= K = ½ m v²
Energy is conserved.
Emo =
m g h = ½ m v²
v² = 2 gh
Let's use trigonometry to find the height, see attached.
h = L - L cos θ
h = L (1- cos θ)
They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.
h = 3.7 (1- cos 48)
h = 1.22 m
this is the maximum height of the movement.
Let's calculate the velocity.
v = 4.89 m / s
In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
Learn more here: brainly.com/question/13010190
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