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Mashcka [7]
3 years ago
12

A 22 kg sled is pushed for 5.2 m with a horizontal force of 20 N, starting from rest. ignore friction. find the final speed of t

he sled.
2.2 m/s
4.7 m/s
9.5 m/s
3.1 m/s
Physics
1 answer:
maksim [4K]3 years ago
5 0
We first calculate the friction on the sled using:
F = ma
a = 20 / 22
a = 0.91 m/s²

Now we use the formula:
2as = v² - u²
v = √(2as + u²)
v = √[2(0.91)(5.2) + 0]
v = 3.08 m/s
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My name is Ann [436]

✯Hello✯

↪  A satellite was crashed into a comet (on purpose of course)

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❤Gianna❤

4 0
3 years ago
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In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro
blondinia [14]
In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass. 
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were. 

Ok, Rameses:

Elementary charge . . . . .  1.6 x 10⁻¹⁹  coulomb
                                        negative on the electron
                                        plussitive on the proton

Electron rest-mass . . . . .  9.11 x 10⁻³¹  kg


a).  The force between two charges is

      F  =  (9 x 10⁹) Q₁ Q₂ / R²

          =  (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

          =     ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

          =          8.05 x 10⁻⁸  Newton .


b).  Centripetal acceleration  = 

                                               v² / r  .

                  A  =  (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)

                     =      7.7 x 10²²  m/s² .

That's an enormous acceleration ... about  7.85 x 10²¹  G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use  "  F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether  F = M A .

I'm going to leave that step for you.   Good luck !
4 0
3 years ago
Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com
Bas_tet [7]

Answer:

Failure rate   = 20%

MTBF = 880 hours

Explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at =  140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷  (number of breakdowns)    ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be  = (Total up time) ÷ (number of breakdowns)      .......2

=  \frac{2000}{2}   =  1000  

so here gap between occurrences is

gap between occurrences=  140 - 20

gap between occurrences = 120 hour

and

MTBF  will be

MTBF = 1000 - 120

MTBF = 880 hours  

and

Failure rate (FR)  will be

Failure rate (FR) =  1 ÷ MTBF    ................3

Failure rate (FR) = R÷T     ......................4

as here R is the number of failures and T is total time

so Failure rate (FR)  = 20%

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