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2 years ago

Склянка с воздухом закрыта «пробкой» из мыльной пены.

1 answer:

5 0

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.

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A cyclist moves at a constant speed of 5 m/s if the cyclist does not accelerate during the next 20 seconds he will travel at?

Verizon [17]

5 m/s because the speed is constant

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3 years ago

ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit

horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

$F = \dfrac{d^4G\delta}{8D^3N}$

$\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm$

$F = \dfrac{d^4G}{8D^3}\times 0.004$

$F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004$

F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

$\tau = \dfrac{8FDk_s}{\pi d^3}$

$\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}$

= 442.6 Mpa

The tensile strength of the steel material of ASTM A229 is equal to 1300 Mpa

now,

$\tau_s \leq 0.45 S_u$

$\tau_s \leq 0.45 \times 1300$

$\tau_s \leq 585\ Mpa$

since corrected stress is less than the $\tau_s$

hence, spring will return to its original shape.

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3 years ago

Read 2 more answers

Curtis, a student in our class, makes the following statement: The puck reached a slightly higher location on the ramp than I pr

Sindrei [870]

Answer and Explanation: No, the explanation is not plausible. The puck sliding on the ice is an example of the Principle of Conservation of Energy, which can be enunciated as "total energy of a system is constant. It can be changed or transferred but the total is always the same".

When a player hit the pluck, it starts to move, gaining kinetic energy (K). As it goes up a ramp, kinetic energy decreases and potential energy (P) increases until it reaches its maximum. When potential energy is maximum, kinetic energy is zero and vice-versa.

So, at the beginning of the movement the puck only has kinetic energy. At the end, it gains potential energy until its maximum.

The representation is as followed:

$K_{i}+P_{i}=K_{f}+P_{f}$

$K_{i}+0=0+P_{f}$

$\frac{1}{2}mv^{2} = mgh$

As we noticed, mass of the object can be cancelled from the equation, making height be:

$h=\frac{v^{2}}{2g}$

So, the height the puck reaches depends on velocity and acceleration due to gravity, not mass of the puck.

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2 years ago

A cart is rolling down a ramp. if the angle of the ramp is increased to make the ramp steeper the cart will

Westkost [7]

The speed of the carts descent will increase? Not quite sure what you’re asking.

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3 years ago

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sergey [27]

The answer is A and B

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AND

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2 years ago