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krok68 [10]
3 years ago
7

what does the term equilibrium refer to? a. the resting position of the wave b. the highest point in the wave c. the lowest poin

t of the wave d. the next subsequent wave e. the period of the wave
Physics
2 answers:
Brums [2.3K]3 years ago
7 0
A. the resting position of a wave
djverab [1.8K]3 years ago
6 0

Answer:

a. the resting position of the wave

Explanation:

The equilibrium refers to the resting position of the wave or the position it would be if it was undisturbed. And this is normally exactly midway between the highest point of the wave and the lowest point of the wave.

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(III) An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable breaks when the e
Iteru [2.4K]

Answer: 12Mg/h

Explanation:

Let the spring is compressed by a distance x,before the lift stops,then

Mg(h+x)= 1/2 kx^2 ............... 1

Kx - Mg = M ( 5g ) ............ 2

Make x the subject in equation 2

Kx = 5Mg + Mg

Kx = 6Mg

x = 6Mg/k ............ 3

Put equation 3 into 1

Mg ( h + x ) = 1/2 kx^2

Mgh + Mgx = 1/2kx^2

Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2

Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2

h =18Mg/k - 6Mg/h

K = 12Mg/h

​

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3 years ago
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How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2
soldier1979 [14.2K]
∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)

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3 years ago
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How much have jellyfish populated since 2015?
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About a mil sience 2014-2015
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Scientists and astronomers have found that in galaxies with central black holes, there are also large star formations near those
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A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car
kenny6666 [7]

Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

x=x_o+v_ot-\frac{1}{2}at^2        (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}

0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

4 0
3 years ago
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