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tankabanditka [31]
2 years ago
11

Two identical light springs with spring constant k3 are now individually hung vertically from the ceiling and attached at each e

nd of a symmetric object, such as a rectangular block with uniform mass density. In this case, with the springs next to each other, we describe them as being in parallel. Find the effective spring constant of the pair of springs as a system in this situation in terms of k3. (Use the following as necessary: k3, M, the mass of the symmetric object, and g, the gravitational acceleration.)
Physics
1 answer:
Alenkasestr [34]2 years ago
7 0

Answer:

 Keq = 2k₃

Explanation:

We can solve this exercise using Newton's second one

                F = m a

Where F is the eleatic force of the spring F = - k x

Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement

          F + F = m a

         k₃ x + k₃ x = m a

         a = 2k₃  x / m

To find the effective force constant, suppose we change this spring to what creates the cuddly displacement

       Keq = 2k₃

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A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle wi
Kamila [148]

Answer:

A.)1.52cm

B.)1.18cm

Explanation:

angular speed of 120 rev/min.

cross sectional area=0.14cm²

mass=12kg

F=120±12ω²r

=120±12(120×2π/60)^2 ×0.50

=828N or 1068N

To calculate the elongation of the wire for lowest and highest point

δ=F/A

= 1068/0.5

δ=2136MPa

'E' which is the modulus of elasticity for alluminium is 70000MPa

δ=ξl=φl/E =2136×50/70000=1.52cm

δ=F/A=828/0.5

=1656MPa

δ=ξl=φl/E

=1656×50/70000=1.18cm

δ=1.18cm

6 0
2 years ago
Read 2 more answers
The weight of spaceman Speff at the surface of planet X, solely due to its gravitational pull, is 389 N. If he moves to a distan
miv72 [106K]

Answer:

mass of the planet X = 5.6 × 10²³ kg.

Explanation:

According to Newtons law of universal gravitation,

F = GM₁M₂/r²

Where F = gravitational force, M₁ = mass of the speff, M₂ = mass of the planet X, G = gravitational constant r = distance between the speff and the planet X

making M₂ The subject of the equation above,

M₂ = Fr²/GM₁ .......................... equation 2

Where F = 24.31 N, r = 1.08×18⁴km ⇒( convert to m ) =1.08 × 10⁴  × 1000 m

r = 1.08  × 10⁷ m, G = 6.67  × 10 ⁻¹¹ Nm²/kg², M₁ = 75 kg

Substituting this values in equation 2,

M₂ = 24.13(1.08  × 10⁷ )²/75( 6.67  × 10 ⁻¹¹)

M₂ = 24.13 × 1.17 × 10¹⁴/500.25 × 10⁻¹¹

M₂ = (28.23 × 10¹⁴)/(500.25 × 10⁻¹¹)

M₂ = 0.056 × 10²⁵

M₂ = 5.6 × 10²³ kg.

Therefore mass of the planet X = 5.6 × 10²³ kg.

8 0
2 years ago
A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f
Snezhnost [94]

Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

\mu mg = m\frac{v^2}{r}

where

\mu is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

v=\sqrt{\mu gr}

And by substituting the data of the problem, we find the speed at which the car begins to skid:

v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s

7 0
2 years ago
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A sound technician is testing the sound acoustics in a theatre for an upcoming music concert. As he moves towards the speakers,
Harlamova29_29 [7]

Answer: Increase in wave frequency

Explanation:

When we talk about acoustics we are dealing with sound waves, and one of their main components along with the velocity and wavelength is the <u>frequency.</u>

In this sense, the frequency of any wave refers to how fast (or slow) a wave oscillates. For example, in the especific case of sound waves when the oscillation is faster, the frequency is higher and the pitch gets higher as well.

6 0
2 years ago
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A 1.00-kilogram ball is dropped from the top of a building. just before striking the ground, the ball's speed is 12.0 meters per
Anarel [89]
During the fall, the potential energy stored in the ball is converted into kinetic energy.
Thus,
PE = KE before hitting the ground
= 1/2 • mv^2
= 1/2 • 1 • 12^2
= 72J
6 0
2 years ago
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