Answer:
The speed of the 270g cart after the collision is 0.68m/s
Explanation:
Mass of air track cart (m1) = 320g
Initial velocity (u1) = 1.25m/s
Mass of stationary cart (m2) = 270g
Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/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
B. third
for every action there is a reaction*
Below are the choices that can be found elsewhere:
a. 268 kJ
<span>b. 271 kJ </span>
<span>c. 9 kJ </span>
<span>d. 6 kJ
</span>
So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have:
<span>Ga (s) --> Ga (g) delta H = 277 kJ/mol </span>
<span>Ga (l) --> Ga (g) delta H = 271 kJ/mol </span>
<span>From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). </span>
<span>Ga (s) --> Ga (l) delta H = 6 kJ/mol </span>
<span>At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). </span>
<span>*ANSWER* </span>
<span>9 kJ/mol (C)</span>
Answer/solution:
Given :
Mass =5kg
T 1 =20 C,T 2 =100 ∘C
ΔT=100−20=80 ∘C
Q=m×C×ΔT
where C= specific heat capacity of water
=4200J/(kgK)
Q=5×4200×80
=1680000 Joule.
=1680KJ
