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
Answer:
(A) 7.9 m/s^{2}
(B) 19 m/s
(C) 91 m
Explanation:
initial velocity (U) = 0 mph = 0 m/s
final velocity (V) = 85 mph = 85 x 0.447 = 38 m/s
initial time (ti) = 0 s
final time (t) = 4.8 s
(A) acceleration = 
= 
= 7.9 m/s^{2}
(B) average velocity = 
=
= 19 m/s
(C) distance travelled (S) = ut + 
= (0 x 4.8) + 
= 91 m
Answer:
A) would be the best explanation because hot air is less dense and would rise to the top of a column of air
Answer:
1.25 m/s^2
Explanation:
m = 76 kg, R = 840 N
Let a be the acceleration of teh elevator in upward direction.
By use of Newton's second law
R - mg = m a
R = m ( g + a)
840 = 76 ( 9.8 + a)
a = 1.25 m/s^2
Answer:
I'm pretty sure it's air pressure.
Explanation:
