<span>D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.</span>
Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
Answer:
a = 0.16
Explanation:
given,
mass of the object 1 = 0.2 kg
mass of the object 2 = 0.3 kg
acceleration when force is on 0.2 kg = 0.4 m/s²
acceleration when both mass are combine = ?
F = m a
F = 0.2 × 0.4
F = 0.08 N
force acting is same and total mass = 0.2 + 0.3 = 0.5 Kg
F = m a


a = 0.16 m/s²
the acceleration acting when both the body is attached is a = 0.16
Answer:
0.5 m
Explanation:
Givens:
ym1 = 2.5 mm
ym2 = 4.5 mm
Ф_1=π / 4
Ф_2=π / 2
We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is
Ym = (ym1 + ym2)cos(Ф_2/2)
By substitution we have
Ym= (0.025 + 0.045)cos(π/4) = 0.496 m
The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore
Ym^2=(ym1^2+ym2^2)
So we have Ym=√0.025^2+0.045^2
= 0.5 m
<h3><u>Answer;</u></h3>
= 1.256 m
<h3><u>Explanation;</u></h3>
We can start by finding the spring constant
F = k*y
Therefore; k = F/y = m*g/y
= 0.40kg*9.8m/s^2/(0.76 - 0.41)
= 11.2 N/m
Energy is conserved
Let A be the maximum displacement
Therefore; 1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2
Thus; A = sqrt((1.20 - 0.55)^2 + m/k*v^2)
= sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)
= 0.846 m
Thus; the length will be 0.41 + 0.846 = 1.256 m