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3 years ago

LETS TALK ABOUT ANYTHING!!! IS IT NORMAL TO ALWAYS BE SAD??? like all the time........

Physics

2 answers:

Gennadij [26K]3 years ago

5 0

Umm, not normal or healthy but common
could be depression

Andru [333]3 years ago

5 0

I think it is I’m never happy lol always thinking about sad stuff always feel sad and put on a fake smile so people won’t ask me if I’m okay or not

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A tin can has a radius of 0.0410m and is 185 m high if the air is removed from inside the can how much forces will the outside a

grin007 [14]

The force exerted on the bottle is 532 N

Explanation:

When the air is removed from inside the bottle, vacuum is created, therefore the external pressure (the atmospheric pressure) is no longer balanced, and it creates a net force downward on the can.

The pressure is related to the force by the equation:

$p=\frac{F}{A}$

where

p is the pressure

F is the force

A is the area of the can

Here we have:

$p=1.01\cdot 10^5 Pa$ is the atmospheric pressure

r = 0.0410 m is the radius of the can, so the area is

$A=\pi r^2 = \pi (0.0410)^2=5.27\cdot 10^{-3}m^2$

And solving for F, we can find the force on the can:

$F=pA=(1.01\cdot 10^5)(5.27\cdot 10^{-3})=532 N$

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

3 0

4 years ago

Calculate the average maximum height for all three trials when the speed of the bottle is 2 m/s, 3 m/s, 4 m/s, 5 m/s, and 6 m/s.

nadya68 [22]

Answer:

idk

Explanation:

5 0

3 years ago

Read 2 more answers

A rescue pilot drops a survival kit while her plane is flying at an ultitude of 2500m with a forward velocity of 95m. If the air

never [62]

Answer:

Approximately $2.1\; \rm km$ , assuming that $g = -9.8\; \rm m \cdot s^{-2}$ .

Explanation:

Let $t$ denote the time required for the package to reach the ground. Let $h(\text{initial})$ and $h(\text{final})$ denote the initial and final height of this package.

$\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})$ .

For this package:

Initial height: $h(\text{initial}) = 2500\; \rm m$ .
Final height: $h(\text{final}) = 0\; \rm m$ (the package would be on the ground.)

Solve for $t$ , the time required for the package to reach the ground after being released.

$\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}$ .

$\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}$ .

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at $95\; \rm m \cdot s^{-1}$ .) From calculations above, the package would travel forward at that speed for about $22.588\; \rm s$ . That corresponds to approximately: $95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km$ .