Answer:
The acceleration is a = 2.75 [m/s^2]
Explanation:
In order to solve this problem we must use kinematics equations.

where:
Vf = final velocity = 13 [m/s]
Vi = initial velocity = 2 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Now replacing:
13 = 2 + (4*a)
(13 - 2) = 4*a
a = 2.75 [m/s^2]
The first choices are correct, because the second choices could happen by things other than light.
Answer:
Explanation:
In the x direction the force will be
½(-w₀)L/2 = -¼w₀L
acting ⅔(L/2) = L/3 below the x axis.
In the y direction the force will be
½(-w₀)L + ½w₀L/2 = -¼w₀L
the magnitude of the resultant will be
F = w₀L √((-¼)² + (-¼)²) = w₀L√⅛
in the direction
θ = arctan(-¼w₀L / -¼w₀L) = 225°
to find the distance, we balance moments
(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]
(√⅛)[d] = ½ [⅔L] + ¼ [⅔L/2] - ¼ [L - ⅓L/2]
(√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]
(√⅛)[d] = ⅓L + ⅟₁₂L - ¼L + ⅟₂₄L
(√⅛)[d] = 5L/24
d = 5L/24 / (√⅛)
d = 5√⅛L/3
Answer:
60,000m
Explanation:
Convert km/h to m/s by multiplying with 1000/3600.
Convert hours to seconds by multiplying with 3600.
Because displacement is a vector quantity and deals with the shortest distance between points, simply plug it into the equation s=vt.
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e