Answer: vf1/vf2= 1/ sqrt(2)
Explanation :on the moon no drag force so we have only the force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics
if the rock travels H to the bottom we can calculate velocity:
vo=0m/s (drops the rock) , yo=0
vf*vf= vo*vo+2g(y-yo)
when the rock is halfway y = H/2 so:
vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)
when the rock reach the bottom y=H so:
vf2*vf2=2*g*H so vf2 = sqrt(2gH)
so vf1/vf2= 1/ sqrt(2)
good luck from colombia
Kinetic energy = (1/2) (mass) (speed)²
BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second. So we'll have to
make that conversion.
KE = (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²
= (725.5) · (48 · 1000 · 1 / 3,600)² (kg) · (km·m·hr / hr·km·sec)²
= (725.5) · ( 40/3 )² · ( kg·m² / sec²)
= 128,978 joules (rounded)
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