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3 years ago

What’s something that is 50cm

Physics

2 answers:

Tom [10]3 years ago

8 0

Answer:

50 centimeters is 1.310 times the height of a Bowling Pin

Firlakuza [10]3 years ago

5 0

Answer:

50 cm is equivalent to 19,6850393701 inches.

Explanation:

A meter has 100 centimeters. 100 millimeters make one centimeter. The centimeter can be written as cm. While calculating the surface area of an object, the unit of measurement becomes cm2.

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Is there a frame of reference one can go into that seems to eliminate gravity as Newton described it?

andriy [413]

Answer:

Yes such a frame exists: a free-fall (free-float frame) frame. This frame of reference is subject only to gravity and no forces such as electromagnetic forces or nuclear forces.

3 0

3 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

$v'^2-u^2=2gh'$

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s$

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

$\Delta t = 2\cdot 0.04 =0.08 s$

8 0

3 years ago

What part of a thunderstorm kills the most people each year?

alexandr1967 [171]

Here is the answer. The part of a thunderstorm that kills the <span> most people each year is the LIGHTNING. Thunder is only the sound created and will not hurt anyone, but it is the lightning that can kill anyone who will be struck by it. Hope this answers your question. Have a great day!</span>

5 0

3 years ago

Read 2 more answers

Based on the data given, in what direction will the car accelerate?

skad [1K]

Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration.
Not up, not down.

Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is

(487-left + 632-right) - (632-right - 487-right) = 145N to the right.

The net force on the car is all to the right.
The car accelerates to the right.

7 0

3 years ago

What is the relationship between temperature and altitude in the stratosphere? (2 points).

stealth61 [152]

As altitude increases, temperature increases. The stratosphere is the part of the atmosphere that starts in the tropopause and ends in the estratopause. In the troposphere, the air is close to the Earth surface. The air surface can absorb more sunlight energy than the air, so the Earth surface heats the air. As you go higher, the distance to the Earth surface is higher, so the temperature is lower. The troposphere ends in the tropopause, where this trend changes. In the estratopause, there is a lot of ozone, which absorbs the dangerous UV radiation and converts into heat. That heat warms the air. So the air which is close to the estratopause is warm because of the heat released by the ozone reactions. The tropopause is far from the Earth surface and far from the ozone layer, that’s why it is cold. So the tropopause is cold and the estratopause is warm, which means: the air becomes warmer <span>as you rise above the tropopause until you get to the estratopause.</span>

8 0

3 years ago