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2 years ago

The total mechanical energy of a simple harmonic oscillating system is

Physics

1 answer:

Hitman42 [59]2 years ago

5 0

Answer:

The total mechanical energy of a simple harmonic oscillating system is Constant

Explanation:

During the movement of an object in simple harmonic motion, the total mechanical energy is constant

The total mechanical energy is given by $E_{Total} = KE +PE = \frac{1}{2} kx^2+\frac{1}{2}mv^2$

By the conservation of energy principle the energy in the simple harmonic motion transforms between kinetic and potential energy while maintaining a constant sum

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A worker kicks a flat object lying on a roof. The object slides up the incline 10.0 m to the apex of the roof, and flies off the

11111nata11111 [884]

Answer:15.20 m

Explanation:

Given

initial velocity $(v_i)=15 m/s$

inclined length=10 m

$\mu _k=0.435$

inclination $\theta =43.5^{\circ}$

$F_{net}=f_r+mg\sin \theta$

$a_{net}=\mu _kg\cos \theta +g\sin \theta$

$a_{net}=0.435\times 9.8\times \cos 43.5+9.8\times \sin 43.5$

$a_{net}=3.09+6.74=9.83 m/s^2$

$v^2-u^2=2as$

$v^2=15^2-2\times (9.83)10$

$v=5.31 m/s$

So Particle launches with a speed of 5.31 m/s at an angle of \theta $=43.5$

$h_{max}=\frac{u^2\sin^2\theta }{2g}$

$h_{max}=\frac{(5.31)^2\sin ^2(43.5)}{2\times 9.81}$

$h_{max}=0.679$

Total height raised is $0.679+\frac{10}{\sin 43.5} =15.20 m$

6 0

2 years ago

Of the three primary forms of subaerial volcanoes, ________ are large cone-shaped mountains that consist of alternating layers o

Alborosie

Answer:

Strato-volcano

Explanation:

Strato-volcanoes are usually characterized by the presence of steep-sided slopes, with distinct craters, and are frequently erupted and conical in appearance. This type of volcano is generally felsic in nature. Due to the presence of high silica content, the magma being highly viscous, moves at a relatively slower rate. These are highly explosive and produce a large number of pyroclastic materials, lava flow, volcanic ashes, and gases.

They are also commonly considered as the composite volcano, and are comprised of alternating tephra and solidified lava layers.

5 0

2 years ago

The seismic activity density of a region is the ratio of the number of earthquakes during a given time span to the land area aff

Natalija [7]

Answer:

0.0059

Explanation:

According to the question the seismic activity density is given by

$\text{Seismic activity density}=\frac{\text{Number of Earthquakes over a given time span}}{\text{The land area affected}}$

Here,

Number of Earthquakes over a given time span = 424

The land area affected = 71300 mi²

So,

$\text{Seismic activity density}=\frac{424}{71300}\\\Rightarrow \text{Seismic activity density}=0.0059$

The seismic activity density is 0.0059

8 0

2 years ago

Read 2 more answers

Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.

aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

6 = (20 / 32.2 + 50 / 32.2)a

2.173a = 6

A = 2.76 ft/s^2

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0

2 years ago

g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the

Svetlanka [38]

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

$R = \frac{u^{2}Sin2\theta }{g}$

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

$R = \frac{u^{2}Sin60 }{9.8}$

R = 0.088u^2

If the angle of projection is 45 degree.

$R = \frac{u^{2}Sin90 }{g}$

R = u^2 / g

5 0

2 years ago