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Masja [62]
4 years ago
14

An empty semi-truck sits on the scale of a weighing station. It is then loaded with 100 irritated pigeons. These two events then

happen in sequence. a) First, the super irritated pigeons just stand on the floor of the truck. b) Second, they take flight inside the truck. They all fly in the enclosed space such that they do not collide with each other or the walls/ceiling/floor. Is the reading on the scale different in a) versus b)? Does the reading on the scale change during the transition of these events?
Physics
1 answer:
mina [271]4 years ago
4 0

Answer:

The reading in the scale is going to be the same, or if it experiences some change it would be minimum. The reason is because the truck will act as a big closed cage, therefore, when the pigeons fly, the air they move with their wings in order to keep flying,  exerts the same force on the closed cage, that if they were standing on the ground.

If the truck however, allows the air flow, the weight might change, because under this scenario, the air flowing could represent less force exerted on the balance.

Explanation:

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A cannon sends a projectile towards a target a distance 1420 m away. The initial velocity makes an angle 35◦ with the horizontal
ss7ja [257]

Answer:

v_{o}=141.51m/s

Explanation:

From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

x=1420m\\g=-9.8m/s^{2}

From the equation of x-position we know that

x=v_{ox}t=v_{o}cos(35)t

Solving for v_{o}

v_{o}=\frac{x}{tcos(35)} =\frac{1420m}{tcos(35)} (1)

Now, if we analyze the equation of y-position we got

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

At the end of the motion y=0

0=v_{o}sin(35)t+\frac{1}{2}gt^{2}

Knowing the equation for v_{o} in (1)

0=\frac{1420}{tcos(35)}tsin(35)-\frac{1}{2}(9.8)t^{2}

\frac{1}{2}(9.8)t^{2}=1420tan(35)

Solving for t

t=\sqrt{\frac{2(1420tan(35))}{9.8} } =14.25s

Now, we can solve (1)

v_{o}=\frac{1420m}{(14.25s)cos(35)}=141.51m/s

6 0
3 years ago
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Why do water and table salt have very different properties?
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Answer:

D)

Explanation:

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How Is budding different from fertilization
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5 0
3 years ago
Which hand is negatively charged?<br> A<br> B<br> C<br> D
vichka [17]

Answer:

the one that has more negative sights so the first hand on your right :)

6 0
3 years ago
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Naturally occurring element X exists in three isotopic forms: X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abund
vova2212 [387]

Answer:

C. 28.09 amu

Explanation:

The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance),  X-29 (28.976 amu, 4.67% abundance) and  X-30 (29.974 amu, 3.10% abundance).

The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.

The atomic weight is computed as follows:

atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of  X-30 × fractional abundance

atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 ×  0.0310

atomic weight = 25.8031871 + 1.3531792 + 0.929194

atomic weight = 28.0855603 amu

To 2 decimal place atomic weight = 28.09 amu

6 0
3 years ago
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