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Mazyrski [523]
3 years ago
14

A student applies 0.3 N of force to a computer mouse. The mouse slides .025 m. How much work did the student do on the mouse?

Mathematics
1 answer:
Kipish [7]3 years ago
8 0

Answer:

  0.0075 N·m

Step-by-step explanation:

Work is the product of force and distance:

  W = fd

  W = (0.3 N)(0.025 m) = 0.0075 N·m

__

<em>Additional comment</em>

1 N·m = 1 J

Use the units you like. You could also write this in terms of millijoules: 7.5 mJ.

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The cost of membership at local country club is $125 per family member per month. Each family also has to pay $75 in service fee
katen-ka-za [31]
To write the equation you start with the 125 because it is monthly and it doesn't say how often you pay the 75 dollars so the equation would be c(f)=125x+75
6 0
3 years ago
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James creates the table shown to represent the function ƒ(x) = x2 + 8x + 12. Determine the zero(s) of the function.
Gennadij [26K]
I believe that the zeros would be -2 and -6, assuming that the first 2 that you wrote in the equation is supposed to be a power to the x.
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Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca
Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

\frac{dQ}{dt} = -rt

Which has the following solution:

Q(t) = Q(0)e^{-rt}

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that Q(6) = (1-0.03)Q(0) = 0.97Q(0). We use this to find r.

Q(t) = Q(0)e^{-rt}

0.97Q(0) = Q(0)e^{-6r}

e^{-6r} = 0.97

\ln{e^{-6r}} = \ln{0.97}

-6r = \ln{0.97}

r = -\frac{\ln{0.97}}{6}

r = 0.0051

So

Q(t) = Q(0)e^{-0.0051t}

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

Q(t) = Q(0)e^{-0.0051t}

0.5Q(0) = Q(0)e^{-0.0051t}

e^{-0.0051t} = 0.5

\ln{e^{-0.0051t}} = \ln{0.5}

-0.0051t = \ln{0.5}

t = -\frac{\ln{0.5}}{0.0051}

t = 135.9

The half-life of the radioactive substance is 135.9 hours.

6 0
3 years ago
During the 2007 baseball season, Wade was up to bat 10 more times than
bearhunter [10]

Answer:

The number of times Ellis get to bat  is  558.

Step-by-step explanation:

Here, let us assume the number of times Ellis bat = m

So, the number of times Dwight bat =  17 times fewer than Ellis

                                                            = m - 17

Also, number of times Wade got to bat  = 10 more times than Dwight

= ( m- 17 )+ 10

Total number of bat times = 1650

So, the number of times ( Wade + Dwight + Ellis) bat together = 1650

⇒   ( m- 17 )+ 10 + (m - 17) + m =  1650

or, 3 m - 34 + 10 =  1650

or, 3  m = 1674

⇒ m = 1674/3 = 558 , or m = 558

Hence, the number of times Ellis get to bat = m = 558.

3 0
4 years ago
Hello my name is juan and I new here are you able to be nice and give me a nice welcome greet
grigory [225]

Answer:

Sure thing mate, I'm hoping you will be satisfied here. Many people here very positive and helpful. I hope you're doing a fine day. Good luck with everything . In addition, in all honestly i your question i just wanted easy question for points. Thanks man .  

Step-by-step explanation:

8 0
4 years ago
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