What is the answer to life and everything in the universe?

Answers

Answer 1

Answer: The answer to life and the universe is 42

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A diverging lens has a focal length of -30.0 cm. An object is placed 18.0 cm in front of this lens.(a) Calculate the image distance.(b) Calculate the magnification.
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A frictionless pendulum is made with a bob of mass 12.6 kg. The bob is held at height = 0.650 meter above the bottom of its trajectory, and then pushedforward with an initial speed of 4.22 m/s. What amount of mechanical energy does the bob have when it reaches the bottom?

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The answer to your question is 55

If the rise and fall of your lungs is considered to be simple harmonic motion, how would you relate the period of the motion to your breathing rate (breaths per minute)? Breaths per minute is an angular frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is its reciprocal. Breaths per minute is an angular frequency. The period is its reciprocal.

Answers

Answer:

Breaths per minute is a frequency. The period is its reciprocal.

Explanation:

In simple harmonic motion, a period (T) is the time taken for one point to start in a position and reach that position again, in other words to complete a cycle or lapse. In this case, a period is the time one takes from starting to inspire the air to releasing all of it from the lungs.

In simple harmonic motion, the frequency (f) is how many times a point completes a cycle or lapse in one unity of time (could be one second, one minute, one hour, etc). In this case, the frequency is how many times one breathes in one minute. This is the breathing rate, since it is breathings per minute. Breaths per minute is a frequency.

Period (T) and frequency (f) relate to each other in the following formulae: $T=(1)/(f)$ or $f=(1)/(T)$ .

Therefore, breaths per minute is a frequency, and since it is related to the period, we say the period is reciprocal to it.

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, causing it to start sliding downward, starting from rest. If it starts 4.25 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 22.0 N?

Answers

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = (92)/(9.8) = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_(net) = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_(net) = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_(net) = \Delta Kinetic \ Energy$

$W_(net) = (1)/(2) m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_(net) = (1)/(2) * m v^2$

Making v the subject

$v = \sqrt{(2 W_(net))/(m) }$

Substituting value

$v = \sqrt{(2 * 309.98)/(9.286) }$

$v =8.17 m/s$

he magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.9 AA , how many turns of wire would you need? Express your answer using two significant figures.

Answers

Answer: 3400

Explanation:

Given

Magnetic field, B = 0.1 T

Diameter of magnet, d = 2 cm = 0.02 m

Length of magnet, l = 8 cm = 0.08 m

Current of the magnet, I = 1.9 A

Number of turns needed, N = ?

To solve this problem, we would use the formula,

N = (LB) / (μI), where

μ = 1.257*10^-6 Tm/A, so that

N = (0.08 * 0.1) / (1.257*10^-6 * 1.9)

N = 0.008 / 2.388*10^-6

N = 3350

N ~ 3400

Therefore, the number of turns of wire needed is 3400

A galilean telescope adjusted for a relaxed eye is 36cm long. If the objective lens has a focal length of 40cm, what is the magnification?

Answers

For this problem, we use the mirror equation which is expressed as:

1/di + 1/f = 1/d0

Magnification is expressed as the ratio of di and d0.

Manipulating the equation, we will have:

M = di/f +1
M = 36/40 + 1
M = 1.9

Hope this answers the equation.

What are supersonic speeds

Answers

I think your answer is speed faster than the speed of sound

Answer:

speeds above 343 m/s

Explanation:

I have taken the test got 100%

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