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Likurg_2 [28]
3 years ago
8

PLEASE HELP ME I FONT HAVE THAT MUCH TIME LEFT!!! THE ALL CAPS ARE TO CATCH YOUR ATTENTION SO NOW THAT I’VE GOT IT, PLEASE HELP

ME!!!

Physics
2 answers:
podryga [215]3 years ago
8 0

Answer:

hmm if it were up to me i would say gravity potential energy and sorry I don't really have a third one hope this helps though.

Alika [10]3 years ago
8 0
I have the same questionnnn
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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%

4 0
3 years ago
A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

#SPJ1

4 0
2 years ago
Which of these is true about kinetic energy but not necessarily true about potential energy
cestrela7 [59]

Kinetic energy is never negative, but potential energy can be.

Potential energy depends on height above some reference level,
and you can pick any level you want as the reference.  So, if the
object is below the reference level you pick, then its potential
energy relative to your reference level is negative. 

What that means is:  You have to lift it / do work on it / give it more
energy than it has now ... in order to move it to the reference level.

(That's exactly the situation with electrons bound to an atom.  Their
energy is considered negative, because we have to do work and
give them more energy to rip them away from the atom.)
_____________________________________

Regarding the other choices:

-- Kinetic energy is scalar ... Yes.  So is potential energy.

-- Kinetic energy increases with height ...
   No. It doesn't, but potential energy does.

-- Kinetic energy depends on position ...
   No. It doesn't, but potential energy does.

3 0
3 years ago
A line _____ on a typical Speed vs. Time graph means an object is experiencing a constant acceleration.
Savatey [412]

Answer:

The answer should be C. slanted upward to the right.

Hope this helps. :-)

7 0
4 years ago
Read 2 more answers
Question 1 of 25
finlep [7]

Answer:

<em>2.753*10^-11N</em>

Explanation:

According to Newton's law of gravitation, the force between the masses is expressed as;

F = GMm/d²

M and m are the distances

d is the distance between the masses

Given

M = 3.71 x 10 kg

m = 1.88 x 10^4 kg

d = 1300m

G = 6.67 x 10-11 Nm²/kg

Substitute into the formula

F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²

F = 46.52*10^(-6)/1.69 * 10^6

F = 27.53 * 10^{-6-6}

F = 27.53*10^{-12}

F = 2.753*10^-11

<em>Hence the gravitational force between the asteroid is 2.753*10^-11N</em>

<em></em>

6 0
3 years ago
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