Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl
1 answer:
Answer:
b) TA = TB = TC
Explanation:
- When the blocks are brought into contact and isolated from the environment, they will exchange heat until they achieve thermal equilibrium.
- During this exchange, the hotter body will lose heat, which will be gained by the cooler body.
- The equilibrium state will be established once this equation is satisfied:
- Substituting the initial temperatures T₀A = 300º C and T₀B = 400ºC, while simplifying for equal block masses mA = mB, enables us to solve for the final temperature, Tfin:
- At equilibrium, when both blocks combine, they will yield a uniform final temperature of 350ºC.
- When block C, also at this temperature, makes contact, all three blocks will simultaneously reflect this final temperature of 350 ºC.
- Therefore, option b) is correct.
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Answer:
Speeds of 1.83 m/s and 6.83 m/s
Explanation:
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where m represents mass,
is the initial speed before impact,
and
are the velocities of the impacted object after the collision and of the originally stationary object after the impact.
Thus,
After the collision, the kinetic energy doubles, therefore:
Substituting the initial velocity of 5 m/s provides the equation needed to proceed.
We know that
leads to
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Answer:
x = v₀ cos θ t, y = y₀ + v₀ sin θ t - ½ g t2
Explanation:
This pertains to a projectile motion scenario. Here, we will express the equations for both the x and y dimensions.
Now, we will apply trigonometry to determine the initial velocity components.
sin θ = / v₀
cos θ = v₀ₓ / v₀
v_{y} = v_{oy} sin θ
v₀ₓ = v₀ cos θ
Next, let's formulate the equations of motion.
X axis
x = v₀ₓ t
x = v₀ cos θ t
vₓ = v₀ cos θ
Y axis
y = y₀ + t - ½ g t2
y = y₀ + v₀ sin θ t - ½ g t2
v_{y} = v₀ - g t
v_{y} = v₀ sin θ - gt
= v_{oy}^2 sin² θ - 2 g y
It is evident that the major distinction lies in the fact that in an inclined launch compared to a horizontal one, the velocity comprises different components