A truck is moving at a constant speed of 50 m/s. A boy riding in the back of the truck throws a newspaper out the back of the tr
1 answer:
You might be interested in
Answer:
B = 191.26 cm
θ = -14.73°
Explanation:
given,
magnitude of the first displacement(A) = 146 cm
at an angle of 124°
resultant magnitude = 137 cm
and angle made with x-axis by the resultant(R) = 32.0°
component of A in X and Y direction
A x = A cos θ = 146 cos 120° = -73 cm
A y = A sin θ = 146 sin 120° = 126.4 cm
now component of resultant in x and y direction
R x = 137 cos 35°
= 112.2 cm
R y = 137 sin 35°
= 78.6 cm
resultant is the sum of two vectors
R = A + B
R x = A x + B x
B x = 112.2 - (-73) = 185.2 cm
B y = R y - A y
B y = 78.6 - 126.4 = -47.8 cm
magnitude of B
B =
B =
B = 191.26 cm
angle
θ = -14.73°
To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,
Here,
v = Velocity
= Linear density (Mass per unit length)
T = Tension
Rearranging to find the Period we have that
As we know that speed is equivalent to displacement in a unit of time, we will have to
Therefore the tension is 5.54N