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victus00 [196]
1 year ago
10

What is the inverse of f ( x ) ? f ( x ) =

la1" title="\frac{2}{x - 6}" alt="\frac{2}{x - 6}" align="absmiddle" class="latex-formula">
Physics
1 answer:
vladimir1956 [14]1 year ago
5 0

The inverse of f(x) is (6x + 2)/x.

<h3>What is an inverse function?</h3>

The inverse function is defined as a function obtained by reversing the given function.

Given that f(x) = 2/(x-6)

To determine the inverse of a function, all you have to do is switch where x and y are and resolve for y.

So after switching x and y,

f(x) = y = 2/(x-6)

becomes

x = 2/(y - 6)

Now, we solve for y regularly.

f(x) = y = 2/(x-6)

Solve for y :

⇒ x(y - 6)  =2

⇒ xy - 6x = 2

⇒ y = (6x + 2)/x

Therefore the inverse of f(x) is (6x + 2)/x

Learn more about inverse function here:

brainly.com/question/2541698

#SPJ1

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A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

4 0
3 years ago
6.For the following questions, use a periodic table and your atomic calculations to find the unknown information about each isot
Mama L [17]
52, mass number equals protons + neutrons
3 0
3 years ago
A subway train accelerates from rest at one station at a rate of 1.30 m/s^2 for half of the distance to the next station, then d
Minchanka [31]

This problems a perfect application for this acceleration formula:

         Distance = (1/2) (acceleration) (time)² .

During the speeding-up half:     1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half:    1,600 meters = (1/2) (1.3 m/s²) T²

Pick either half, and divide each side by  0.65 m/s²: 

                         T² = (1600 m) / (0.65 m/s²)

                         T = square root of (1600 / 0.65) seconds

Time for the total trip between the stations is double that time.

                         T =  2 √(1600/0.65) = <em>99.2 seconds</em>  (rounded)


7 0
3 years ago
A 20 kg bike accelerates at 10 m/s2. What was the force?
agasfer [191]

Answer:

200 N = 200 Newtons

Explanation:

Just use the formula F = m*a

F = Force in Newtons

m = mass and is 20 kg

a = acceleration and is 10 m/s^2

F = 20 * 10

F = 200 Newtons.

7 0
3 years ago
The average specific heat of the human body is 3.6 kJ/kg·°C. If the body temperature of a(n) 96-kg man rises from 37°C to 39°C d
d1i1m1o1n [39]

Answer:

691200 J

Explanation:

From specific heat capacity,

ΔQ = cmΔt.................. Equation 1

Where ΔQ = increase in thermal energy, c = specific heat capacity of the body, m = mass of the man, Δt = rise in temperature.

Given: c = 3.6 kJ/kg.°C = 3600 J/kg.°C, m = 96 kg, Δt = 39-37 = 2 °C.

Substitute into equation 1

ΔQ = 3600×96×2

ΔQ = 691200 J.

Hence the change in the thermal energy of the body = 691200 J

5 0
3 years ago
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