Answer:
24.084 m/s
Explanation:
From the law of conservation of linear momentum
Total momentum before collision equals to the total momentum after collision
Since momentum=mv where m is mass and v is velocity
where
is the mass of the truck,
is velocity of the truck,
is the common velocity of moving and standing truck after collision and
is the mass of the standing truck
Making
the subject we obtain
Substituting
as 25000 Kg,
as 22.3 m/s,
as 2000 Kg we obtain
Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive
The truck was moving at 24.084 m/s
Answer: 2.7 m/s
Explanation:
Given the following :
Period (T) = 8.2 seconds
Radius = 3.5 m
The tangential speed is given as:
V = Radius × ω
ω = angular speed = (2 × pi) / T
ω = (2 × 22/7) / 8.2
ω = 6.2857142 / 8.2
ω = 0.7665505
Therefore, tangential speed (V) equals;
r × ω
3.5 × 0.7665505 = 2.6829268 m/s
2.7 m/s
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
I think the answer is CuF2