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Sholpan [36]
3 years ago
8

Which layer of the atmosphere is just above the troposphere?

Physics
1 answer:
boyakko [2]3 years ago
6 0
The answer would Stratosphere because <span>1) The </span>troposphere<span> is the first </span>layer above<span> the surface and contains half of the Earth's </span>atmosphere<span>. Weather occurs in this </span>layer. 2) Many jet aircrafts fly in thestratosphere<span> because it is very stable. Also, the ozone </span>layer<span> absorbs harmful rays from the Sun</span>
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A scale contains a spring with a spring constant of 291 N/m. Placing a mass on the scale causes the spring to be compressed by 2
WINSTONCH [101]
E = 1/2*k*x^2 = 0.5*291*0.0289^2 = 0.12 J
6 0
3 years ago
If the resistance of a circuit increases, that means that the current has to
Gnoma [55]

Answer:

If resistance increases current decreases.

Explanation:

  • Current is <em>inversely proportional</em> to the resistance.
  • from the relation given below, we can clearly see the relation between current and resistance;

                              V=IR

                              I ∝ 1/R

This relation shows that when resistance increases,current decreases.

4 0
3 years ago
The height of the Washington Monument is measured to be 170 m on a day when its temperature is 35.0°C. What will the change in i
Alecsey [184]

Answer:

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m

Explanation:

Thermal coefficient of marble varies between (5.5 - 14.1) ×10⁻⁶/K = α

So, let us take the average value

(5.5+14.1)/2 = 9.8×10⁻⁶ /K

Change in temperature = 35-(-18) = 53 K = ΔT

Original length = 170 m = L

Linear thermal expansion

\frac{\Delta L}{L} = \alpha\Delta T\\\Rightarrow \Delta L=\frac{\alpha\Delta T}{L}\\\Rightarrow \Delta L=9.8\times 10^{-6}\times 53\times 170

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m (subtraction because of cooling)

4 0
3 years ago
An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr
notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

       Charge = Current \times Time

As two different current is passing at two different times, the net charge will be the different in current.  So,

        \text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

        V=\frac{k q}{R}

Here k = 9 * 10^9, q is the charge and R is the radius. As q=2 \times 10^{-6} \times t and R =17 cm = 0.17 m, then the voltage will be

        V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

The time is required to find to reach the voltage of 1500 V, so

1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}

So, 14 ms is required to reach the potential of 1500 V.

3 0
3 years ago
It is correct to say that impulse is equal toA) momentum.B) the change in momentum.C) the force multiplied by the distance the f
Elena-2011 [213]

Answer:

B) the change in momentum.

Explanation:

The impulse is defined as the product between the force applied on an object (F) and the duration of the collision (\Delta t):

J=F \Delta t (1)

We can rewrite the force by using Newton's second law, as the product between mass (m) and acceleration (a):

F=ma

So, (1) becomes

J=ma \Delta t

Now we can also rewrite the acceleration as ratio between the change in velocity and change in time: a=\frac{\Delta v}{\Delta t}. If we substitute into the previous equation, we find

J=m\frac{\Delta v}{\Delta t}\Delta t=m\Delta v

And the quantity m\Delta v is equivalent to the change in momentum, \Delta p.

6 0
3 years ago
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