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
Vi = 15 m/s
t = 2 s
a = 9.8 m/s^2
y = ?
The kinematic equation that has all of our variables is d = Vi*t + 0.5*a*t^2
y = 15*2 + 0.5*9.8*2^2 = 49.6 m
Answer:
Explanation:
given
m= 17.5kg
F= 75N
d= 5.7m
∪=0.150
θ= 21°
a. W = Fcos θ × d
75cos21° ×5.7
=399.106J
b. normal force is zero. 0 Joules
cos 90°=0
Answer:
exothermic
Explanation:
energy is absorbed by the surroundings
Answer:
Explanation:
a ) period of motion = 60 / 5 = 12 s
b ) At the highest level
centripetal acceleration will always be towards the centre so it will be vertically downwards
magnitude
= ω²R
= (2π / T )² R
= (4π² / T² )x R
= (4π² / 12²)x 15
= 4.1 m s⁻²
b ) At the lowest level
centripetal acceleration will always be towards the centre so it will be vertically upwards
Magnitude
Magnitude will be same because woman is moving with uniform speed and radius is also same.
