Answer:
Fg = 98.1 [N]; N = 98.1 [N]; Ff = 39.24 [N]; a = 2.076[m/^2]
Explanation:
To solve this problem, we must make a free body diagram and interpret each of the forces acting on the box. In the attached diagram we can find the free body diagram.
The gravitational force is equal to:
Fg = (10 * 9.81) = 98.1 [N]
Now by summing forces on the Y axis equal to zero, we can find the normal force exerted by the surface.
N - Fg = 0
N = Fg
N = 98.1 [N]
The friction force is defined as the product of normal force by the coefficient of friction.
Ff = N * μ
Ff = 98.1 * 0.4
Ff = 39.24 [N]
By the sum forces on the x-axis equal to the product of mass by acceleration (newton's second law), we can find the value of acceleration.
60 - Ff = m * a
60 - 39.24 = 10 * a
a = 2.076[m/^2]
Answer:
y = 33.93 10⁵ m
Explanation:
This is an interference exercise, for the contributory interference is described by the expression
d sin θ = m λ
let's use trigonometry for the angle
tan θ = y / L
how the angles are small
tan θ = sin θ / cos tea = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
the light fulfills the relation of the waves
c = λ f
λ = c / f
λ = 3 10⁸ /375
λ = 8 10⁵ m
first order m = 1
let's calculate
y = 1 8 10⁵ 4030 10-9 / 950 10-9
y = 33.93 10⁵ m
Answer:
The magnitude of the tension on the ends of the clothesline is 41.85 N.
Explanation:
Given that,
Poles = 2
Distance = 16 m
Mass = 3 kg
Sags distance = 3 m
We need to calculate the angle made with vertical by mass
Using formula of angle



We need to calculate the magnitude of the tension on the ends of the clothesline
Using formula of tension

Put the value into the formula


Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.
Answer:
The answer is 100J.
Explanation:
In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. In this question, the mass is equals to 50kg and the velocity is 2m/s
Now,
25kg×4m/s^2 = 100kgm/s^2 or 100J