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rusak2 [61]
3 years ago
14

In the formula used to solve problems related to the first law of thermodynamics what does Q represent?

Physics
1 answer:
KonstantinChe [14]3 years ago
4 0
The heat <span>Q(in)</span> supplied to the system in one stage of the cycle, minus the heat <span>Q(out)</span> removed from it in another stage of the cycle; plus the work added to the system <span>W(in)</span> equals the work that leaves the system <span>W(<span>out)</span></span>
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A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity
Viktor [21]

Answer:

-26 m/s.

Explanation:

Hello,

In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:

t=\frac{0m/s-26m/s}{-9.8m/s^2} =2.65s

With which we compute the maximum height:

y=26m/s*2.65s-\frac{1}{2}*9.8m/s^2*(2.65s)^2 \\\\y=34.5m

Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:

v_f=\sqrt{0m/s-(-9.8m/s^2)*2*34.5m}\\ \\v_f=-26m/s

Which is clearly negative since it the projectile is moving downwards the starting point.

Regards.

3 0
3 years ago
What are two reasons why a home computer scanner requires
alexandr1967 [171]

Answer: C and D

Explanation: a p e x

4 0
2 years ago
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What is the motion of the particles in this kind of wave?
Musya8 [376]
The answer to this question is B I think
7 0
3 years ago
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A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi
Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L)  ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L

Finally, using the Parallel Axis Theorem, we calculate I_B:

I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4}  ML^{2} =\frac{1}{3} ML^{2}

5 0
3 years ago
Robin would like to shoot an orange in a tree with his bow and arrow. The orange is hanging yf=5.00 myf=5.00 m above the ground.
tensa zangetsu [6.8K]

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

<u>h' = 55.3 m</u>

4 0
2 years ago
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