Склянка с воздухом закрыта «пробкой» из мыльной пены.
Определи, что произойдёт, если склянку частично погрузить в ёмкость с горячей водой?
Выбери правильный вариант ответа.
Мыльная пена будет перемещаться внутрь склянки.
Мыльная пена будет подниматься наружу.
Что произойдёт с давлением воздуха внутри склянки?
Выбери правильный вариант ответа.
Давление не изменится.
и дальнейшее объяснение произошедшего
Может показаться, что воздух в вашей бутылке - просто пустое пространство, но на самом деле это смесь газов. Сила, с которой отдельные молекулы отскакивают от внутренней и внешней стороны бутылки, называется давлением.
Explanation:
An Example of push as a force would be to push on a swing. The force moves the swing in a particular direction and the harder that you push the further the swing will go.
An example of pull as a force would be opening a door. ...
An example of pressure as a force is when you push down on a pile of grapes. is this what you mean
Answer:
d 
= m× λ⇒ d = λ ×m×l / x
= 630×
m × 3×3m/ 45×
m
= 1.26×
m
Explanation:
the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.
I am sitting in my seat.
I am listening to my mp3 and reading my book.
My eyes are getting heavy. They start to close.
I try to stay awake, but it's no use.
I am so warm and comfortable and sleepy,
and I have just finished my dinner.
Finally I can't help it. Resistance is futile.
I give up, and fall deep asleep.
My head rests back against my soft, comfy seat.
My seat is in row 26 on the airplane I'm flying in
to visit my grandmother on the coast.
We are cruising at 560 miles an hour, bearing 280°,
at flight level 320 .
The temperature outside my window is -60°F .
Answer:
T1 = 130N, T2 = 370N
Explanation:
In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.
1. forces:
Let tension in rope 1 be T1 and in rope 2 be T2:
ma = T1 + T2 - 100N - 400N = 0
(1) T1 + T2 = 500N
2. torque around the center point of the beam:
τ = r x F = 5*T1 + 3*400N - 5*T2 = 0
(2) T1 - T2 = -240N
Solving both equations:
T1 = 130N
T2 = 370N
