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4 years ago

10

a mass stands on a platform which vibrates simple harmonically in a vertical Direction and with a frequency of 5 Hertz. show tha

t the mass loses contact with the platform when the displacement exceeds 10^-2.

1 answer:

Nastasia [14]4 years ago

4 0

shm

displacement is A*sin 2pi f t

vel =( A*2*pi*f*)*cos 2 pi f t

accn =-( A*2*pi*f*)^2*sin 2 pi f t

accn = (0.01*2*3*5)

accn = (0.01x30)^2=9

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Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous

Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is Av with A the cross section area and v the velocity,

so,

$A_1V_1=A_2V_2$

$A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}$

the diameter decreases 86% so

$d_2 = 0.86d_1$

$v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}$

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3 0

3 years ago

How does the resulting cell at the end of asexual reproduction compare to the original cell

lina2011 [118]

1. At the end of asexual reproduction the resulting cell is an exact genetic copy of the original cell.

2. During inter phase the cell copies its DNA, and grows to an appropriate size where it is safe for the cell to split.

3. After meiosis, the resulting daughter cells are 4 genetically different haploid daughter cells.

4. During sexual reproduction, two haploid gametes join in the process of fertilization to produce a diploid zygote. The two haploid gamete bring different traits together, they also shares some DNA. This makes each daughter cell unique.

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3 years ago

A ballast is dropped from a stationary hot-air balloon that is at an altitude of 576 ft. Find (a) an expression for the altitude

Nadya [2.5K]

Answer:

<h2>a) $S = \frac{1}{2}gt^2\\$ </h2><h2>b) 6secs</h2><h2>c) 192ft</h2>

Explanation:

If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;

$S = ut + \frac{1}{2}at^{2}$

S is the altitude of the ballest

u is the initial velocity

a is the acceleration of the body

t is the time taken to strike the ground

Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s

Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g

Substituting this values into the equation of the motion;

$S = 0 + \frac{1}{2}gt^2\\ S = \frac{1}{2}gt^2\\$

a) An expression for the altitude of the ballast after t seconds is therefore

$S = \frac{1}{2}gt^2\\$

b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);

$576 = \frac{1}{2}(32)t^2\\\\\\576*2 = 32t^2\\1152 = 32t^2\\t^2 = \frac{1152}{32} \\t^2 = 36\\t = \sqrt{36}\\ t = 6.0secs$

This means that the ballast strikes the ground after 6secs

c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.

v = 0 + 32(6)

v = 192ft

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4 years ago

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yanalaym [24]

Answer:

Work done by the spring is negative

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We can answer this question by thinking what is the force acting on the box.

In fact, the force acting on the box is the restoring force of the spring, which is given by Hooke's Law:

$F=-kx$

where

k is the spring constant

x is the displacement of the box with respect to the equilibrium position of the spring

The negative sign in the equation indicates that the direction of the force is always opposite to the direction of the displacement: so, whether the spring is compressed or stretched, the force applied by the spring on the box is towards the equilibrium position.

The work done by the restoring force is also given by

$W=Fx cos \theta$

where

F is the restoring force

x is the displacement

$\theta$ is the angle between the direction of the force and the displacement

Here we know that the force is always opposite to the displacement, so

$\theta=180^{\circ} \rightarrow cos \theta =-1$

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3 years ago

Formulating an Investigative Question

valentina_108 [34]

Answer:

How do mass and speed affect kinetic energy?

Explanation:

This was edges sample resopnse so do not copy and just put this in your own words.

Have a great day whoever reads this.

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3 years ago