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kolbaska11 [484]
3 years ago
13

When flying in an airplane, you are most likely in which layer of the atmosphere? mesosphere thermosphere stratosphere trosphere

Physics
2 answers:
Sergeeva-Olga [200]3 years ago
6 0
Lower stratosphere, this is to avoid turbulence
Oliga [24]3 years ago
3 0

Answer:

its troposphere!!

Explanation:

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How does a comet change as it travels through space
AfilCa [17]
Comets are like "dirty snowballs"; frozen gasses with dust and rocks in them. Each pass near the Sun causes the comet's nucleus to be exposed to intense sunlight, which causes some tiny fraction of the gas to evaporate and carry some of the dust and rock away into space. The gas and dust, near the Sun, cause the comet's "tail", and repeated passes cause dust and rock to spread out along most of the orbit of a comet. When the Earth enters one of these trails of old comet dust, we have meteor showers. 

<span>On rare occasions, comets break apart or even more rarely, crash into planets. In 1994, the comet Shoemaker-Levy 9 broke apart and then collided with the planet Jupiter.</span>
5 0
3 years ago
A 0.5 μF and a 11 μF capacitors are connected in series. Then the pair are connected in parallel with a 1.5 μF capacitor. What i
NISA [10]

Answer:

C_{eq}=1.97\ \mu F

Explanation:

Given that,

Capacitance 1, C_1=0.5\ \mu F

Capacitance 2, C_2=11\ \mu F

Capacitance 3, C_3=1.5\ \mu F

C₁ and C₂ are connected in series. Their equivalent is given by :

\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}

\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}

C'=0.47\ \mu F

Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :

C_{eq}=C'+C_3

C_{eq}=0.47+1.5

C_{eq}=1.97\ \mu F

So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.

3 0
3 years ago
Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

7 0
3 years ago
Three point charges are fixed in place in a right triangle, as shown in the figure.
8090 [49]
Oh gosh oh I see it in my life face and
4 0
3 years ago
A homing pigeon starts from rest and accelerates uniformly at +4.00 m/s squared for 10.0 seconds. What is its velocity after the
zavuch27 [327]
'  +4 m/s² ' means that the pigeon's speed is  4 m/s greater every second.

Starting from zero speed, after 10 seconds, its speed is 

                           (10 x 4m/s)  =  40 m/s.

We can't say anything about its velocity, because we have
no information regarding the direction of its flight.
5 0
3 years ago
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