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Marianna [84]
3 years ago
13

Explain why sublimation and deposition are classified as physical changes

Physics
1 answer:
Over [174]3 years ago
7 0
Because even though it goes through a phase change it's still water
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Kinetic energy depends on the height of an object above the ground?
s344n2d4d5 [400]
Annnnd? Ummmmmmmmmmmmmmm
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3 years ago
Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera
torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

  • No intersection: There's nothing that blocks the camera's view of the top of the building.
  • Two intersections: The planet blocks the camera's view of the top of the building.
  • One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle \angle \mathrm{B\hat{C}D} which corresponds to this minor arc.

This angle comes can be split into two parts:

\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}.

Also,

\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}.

The radius of this circle is:

\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}.

The lengths of segment DC, AC, BC can all be found:

  • \rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m;
  • \rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m;
  • \rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m.

In the two right triangles \triangle\mathrm{DAC} and \triangle \rm BAC, the value of \angle \mathrm{B\hat{C}A} and \angle \mathrm{A\hat{C}D} can be found using the inverse cosine function:

\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}

\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}

\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}.

The length of the minor arc will be:

\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km.

5 0
3 years ago
A hammer of mass m falls from rest off of a roof and drops a height H onto your head.
belka [17]
For conservation of energy we have to:
 mgH=mv²/2 
 Clearing
<span> v=sqrt(2gH)
 Then, by definition 
</span><span> F=Δp/Δt= Δ(mv)/ Δt=m Δ(v)/Δt= 
</span> =m[sqrt(2gH)-0]/Δt= m[sqrt(2gH)]/ Δt
 the answer is
 F=m[sqrt(2gH)]/ Δt
 
6 0
3 years ago
I need c only please answer it
Elodia [21]
It will be moving at high speeds
5 0
3 years ago
lenny accidentally drops his matchbox car into a graduated cylinder. if the car has a mass of 10 grams and the water level in th
Rina8888 [55]
Density formula is:
D = m / V
Volume of the matchbox car:
V = 46 ml - 41 ml = 5 ml = 0.005 l = 0.005 dm³
D = 10 g / 0.005 dm³
D = 2000 g / dm³ = 2 kg / dm³ = 2 × 10 ^(-3) kg/m³
4 0
3 years ago
Read 2 more answers
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