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3 years ago

How is it that even light precipitation can still cause the collection of large amounts of water.

Physics

1 answer:

Firdavs [7]3 years ago

4 0

Answer:

the heat of the light.

Explanation:

no matter the light, there's always heat being produced from it. and heat makes liuqid rise

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5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t

Volgvan

Answer:

When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.

5 0

4 years ago

If it takes 726 watts of power to move a mass 36 meters in 14 seconds, then what is the magnitude of the object’s mass?

Colt1911 [192]

Answer:

20.2

Explanation: Divide 726/34 and you get 20.16 round it and you get 20.2

4 0

3 years ago

The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulu

Gemiola [76]

Answer:

$\displaystyle w=3.478\ rad/sec$

$M=0.0182\ J$

$v=0.398\ m/s$

Explanation:

<u>Simple Pendulum</u>

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

(a) The angular frequency of the motion is computed as

$\displaystyle w=\sqrt{\frac{g}{L}}$

We have the length of the pendulum is L=0.81 meters, then we have

$\displaystyle w=\sqrt{\frac{9.8}{0.81}}$

$\displaystyle w=3.478\ rad/sec$

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

$U=mgh$

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

$H+Y=L$

And

$H=L\ cos\alpha$

Solving for Y

$Y=L(1-cos\alpha )$

$Since\ \alpha=8.1^o, L=0.81\ m$

$Y=0.0081\ m$

The potential energy is

$U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)$

$U=0.0182\ J$

The mechanical energy is, then

$M=K+U=0+U=U$

$M=0.0182\ J$

(c) The maximum speed is achieved when it passes through the lowest point (the reference for h=0), so the mechanical energy becomes all kinetic energy (K). We know

$\displaystyle K=\frac{mv^2}{2}$

Equating to the mechanical energy of the system (M)

$\displaystyle \frac{mv^2}{2}=0.0182$

Solving for v

$\displaystyle v=\sqrt{\frac{(2)(0.0182)}{0.23}}$

$v=0.398\ m/s$

4 0

3 years ago

A ladder 7.85 m long leans against the side of a building. If the ladder is inclined at an angle of 79.5° to the horizontal, wha

Slav-nsk [51]

Jsdjdjjjdjdjd

Explanation:Sjsjsjsmjsjddhdmddjdidujddjdujddmjsisisi daddy dickkkdududjdjsjsusjsususuksuckmpormmjsjdjsjxjxxjjjsusjhs

5 0

4 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

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