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tamaranim1 [39]
2 years ago
10

For elevator questions, does the weight of a person increase when the elevator is going up or if it's going down-

Physics
1 answer:
TiliK225 [7]2 years ago
6 0

The weight of a person increase when the elevator is going up.

<h3>Weight of the person in the elevator</h3>

The weight of the person in the elevator is calculated as follows;

<h3>When the person is going up</h3>

F = ma + mg

F = m(a + g)

where;

  • a is acceleration of the person
  • g is acceleration due to gravity

<h3>When the person is going down</h3>

F = mg - ma

F = m(g - a)

Thus, the weight of a person increase when the elevator is going up.

Learn more about weight here: brainly.com/question/2337612

#SPJ1

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A car traveling at 30 m/s speeds up to 35 m/s over a period of 5 seconds. What is the acceleration of the car?
Delvig [45]

Answer:

u =30 m/s

v = 35 m/s

t = 5 secs

Explanation:

a = (v- u)/t

a = (35-30)/5

a = 5/5

a = 1 m/s^2

PLS MARK BRAINLIEST

8 0
2 years ago
An electric current in a metal consists of moving
Alik [6]
The answer would be (A) Protons
6 0
3 years ago
when a hockey stick strike a hockey puck and sends it towards the goalie, (thermal, mechanical) energy is being transferred from
velikii [3]

Answer:

Yes... definitely true.. can you tell me what is the question..

Explanation:

Please mark me brainliest :D

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8 0
2 years ago
please help In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ba
viva [34]

Answer:

the acceleration during the collision is: - 5  \frac{m}{s^2}

Explanation:

Using the formula:

a=\frac{\Delta\,v}{\Delta\,t}

we get:

a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}

4 0
4 years ago
If a star with an absolute magnitude of -5 has an apparent magnitude of +5 ,then its distance is
klio [65]
You asked a question.  I'm about to answer it. 
Sadly, I can almost guarantee that you won't understand the solution. 
This realization grieves me, but there is little I can do to change it. 
My explanation will be the best of which I'm capable.


Here are the Physics facts I'll use in the solution:

-- "Apparent magnitude" means how bright the star appears to us.

-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).

-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by  ⁵√100 .
That's about  2.512... .  

-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .

-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).

That's all the Physics.  The rest of the solution is just arithmetic.
____________________________________________________

-- The star in the question would appear M(-5) at a distance of
32.6 light years. 

-- It actually appears as a M(+5).  That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.

-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .

-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is

(32.6) · (100)^(1) light years

= (32.6) · (100) light years

=  approx.  3,260 light years .   (roughly 1,000 parsecs)


I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
6 0
3 years ago
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