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3 years ago

What type of cloud is highest in the sky?

Physics

2 answers:

HACTEHA [7]3 years ago

4 0

Cirrus clouds are the highest in they sky

Hope this helps!

Serga [27]3 years ago

4 0

Cirrus is the answer.

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A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the stri

frutty [35]

Answer:

-2.26×10^-4 radians

Explanation:

The solution involves a right angle triangle

Length is z while the horizontal is the height x

X^2+ 100^2=z^2

Taking the derivatives

2x(dx/dt)=Z^2(dz/dt)

Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11

dz/dt= 1100sqrt3/200 = 9.53

Sin a= 100/a

Taking derivatives in terms of t

Cos a(da/dt)=100/z^2 dz/dt

a= 30°

Cos (30°)da/dt= (-100/40000×9.5)

a= -2.26×10^-4radians

8 0

3 years ago

Read 2 more answers

A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a sy

Akimi4 [234]

Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be $F_1$ and force on +Q charge y axis due to -Q charge on x-axis be $F_2$ .

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = $\sqrt{2}d$ units

$k_e$ = Coulomb constant

$F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N$

$F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N$

Net force on +Q charge on y-axis is:

$F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N$

$F_y=F_1-F_2cos45^o$

$F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})$

$F_N=\sqrt{F_x^2+F_y^2}$

$|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|$

The net froce on the +Q charge on y-axis is

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

4 0

3 years ago

Read 2 more answers

Can someone help?

laila [671]

Answer:

i) 21 cm

ii) At infinity behind the lens.

iii) A virtual, upright, enlarged image behind the object

Explanation:

First identify,

object distance (u) = 42 cm (distance between object and lens, 50 cm - 8 cm)

image distance (v) = 42 cm (distance between image and lens, 92 cm - 50 cm)

The lens formula,

$\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

Then applying the new Cartesian sign convention to it,

$\frac{1}{v} +\frac{1}{u} =\frac{1}{f}$

Where f is (-), u is (+) and v is (-) in all 3 cases. (If not values with signs have to considered, this method that need will not arise)

Substituting values you get,

i) $\frac{1}{42} +\frac{1}{42} =\frac{1}{f}\\\frac{2}{42} =\frac{1}{f}$

f = 21 cm

ii) u =21 cm, f = 21 cm v = ?

Substituting in same equation $\frac{1}{v} =\frac{1}{21} =\frac{1}{21} \\\\\frac{1}{v} = 0\\$

v ⇒ ∞ and image will form behind the lens

iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.

7 0

3 years ago

You are given two temperatures for substance: one is the melting point and one is the boiling point. How do you determine which

Sergio039 [100]

melting point is where the substance is melting and turning from a solid to a liquid.

boiling point is where molecules vibrate and produce energy which makes the water hot and becomes a boil.

8 0

3 years ago

What would cause gravity’s force and why would it?

pickupchik [31]

Answer:

Earth's gravity comes from all its mass. All its mass makes a combined gravitational pull on all the mass in your body. ... You exert the same gravitational force on Earth that it does on you. But because Earth is so much more massive than you, your force doesn't really have an effect on our planet.

Explanation:

8 0

2 years ago