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Artist 52 [7]
3 years ago
15

In which of the following elevator situations would the acceleration be positive. Select TWO answers

Physics
1 answer:
kap26 [50]3 years ago
4 0

Both options 5 and 6

Explanation:

Let us consider option 5,

In option 5 body is moving up with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

Let us consider option 6,

In option 6 body is moving down with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

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BRAINLIEST+10PTS: does ozone absorb just harmful UV rays? Or just a random portion of UV?
emmasim [6.3K]

Answer:As sunlight passes through the atmosphere, all UVC and most UVB is absorbed by ozone, water vapour, oxygen and carbon dioxide. UVA is not filtered as significantly by the atmosphere. Is there a connection between ozone depletion and UV radiation? Ozone is a particularly effective absorber of UV radiation.

Explanation:

I had old notes

4 0
2 years ago
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A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of ki
earnstyle [38]

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero. 

Since you're pushing on it with 53N in <em><u>that</u></em> direction, friction must be pulling
on it with 53N in the <u><em>other</em></u> direction.

 The 53N of friction is (the weight) x (the coefficient of kinetic friction).

                                                  53N  =  (490N) x (coefficient).

Divide each side by  490N :  Coefficient = (53N) / (490N)  =  0.1082 .

Rounded to the nearest hundredth, that's    <em>0.11 </em>.      (choice 'd')


5 0
3 years ago
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A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the
Marrrta [24]

Our values can be defined like this,

m = 65kg

v = 3.5m / s

d = 0.55m

The problem can be solved for part A, through the Work Theorem that says the following,

W = \Delta KE

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

\frac {1} {2} mv ^ 2

So,

Fd = \frac {1} {2} m ^ 2

Clearing F,

F = \frac {mv ^ 2} {2d}

Replacing the values

F = \frac {(65) (3.5)} {2 * 0.55}

F = 723.9N

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0
3 years ago
An airtight box has a removable lid of area 1.10 10-2 m2 and negligible weight. The box is taken up a mountain where the air pre
andrezito [222]

Answer:

F=7.7\times10^2 N

Explanation:

The magnitude of force required to pull the lid off the box by air pressure.

We know that Pressure, P= Force(F)/Area(A)

Force, F= P×A

Given: A=1.10\times10^{-2} m^2

P=7\times10^{4} Pa.

Therefore, F=7\times10^{4}\times1.10\times10^{-2}.

F=7.7\times10^2 N

4 0
4 years ago
Un joven pelotero llamado Saúl en su primer año de ser firmado en grandes ligas, batea varias veces de cuadrangular, el segundo
Brilliant_brown [7]

Answer:

135-15=120

120÷3=40

40+40+(10)+40+(5)=135

8 0
3 years ago
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