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Setler79 [48]
4 years ago
14

Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is

their effective spring constant? 004393 Submit Answer Incorrect. Tries 4/99 Previous Tres If a mass of 0.31 kg is attached what will be the frequency of oscillation?
Physics
1 answer:
Llana [10]4 years ago
4 0

Answer:

K_e_q=22.75878093\frac{N}{m}

f=1.363684118Hz

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}

Replacing the data provided:

\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}

K_e_q=22.75878093\frac{N}{m}

Finally, to calculate the frequency of oscillation we use this:

f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }

Replacing m and k:

f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz

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A force acts on a body of mass 13 kg initially at rest. The force
PtichkaEL [24]

Answer:

Force that acted on the body was F = 13 N

Explanation:

If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s

When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:

v_f=v_i+a*t

which in our case becomes;

10\,m/s=0+a*(10\,s)

and we can solve for the acceleration as:

a = 10/10  m/s^2 = 1 m/s^2

Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:

F = 13 kg * 1 m/s^2 = 13 N

4 0
3 years ago
A spectroscope creates a spectrum, or array of colors, based on the light emitted by a star. How are
Ivanshal [37]
In a spectrograph, black lines can be seen going through the array of colors. The pattern of these lines indicate the composition of the star.

Different elements block different parts of the spectrum, resulting in black lines.

6 0
3 years ago
A bicycle wheel rotates counterclockwise at a constant rate. You can use one of the wheel’s spokes to measure the wheel’s angula
nevsk [136]

Answer: s=delta/360° ×2πr

Explanation:

The center of the wheel is located.The distance from the centre spoke to the outside wheel is the radius r...the distance between two spokes in the wheels is an arc which forms an angle(theta) at the centre... It is a fraction of the angle formed at the middle to the total angle of the wheel(360° as in circle) to the product of the circumference of the wheel(circle)

Length of arc(s)=theta/360 ×2πr

4 0
3 years ago
If a material, such as concrete, can withstand a large applied weight but gives away (ruptures) quickly when a crack forms, then
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3 years ago
Read 2 more answers
In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm,
Serga [27]

Answer:

4E

Explanation:

From the question given above, the following data were obtained:

Initial elongation (e₁) = 4 cm = 4/100 = 0.04 m

Initial energy (E₁) = E

Final elongation (e₂) = 0.04 + 0.04 = 0.08 m

Final energy (E₂) =?

The energy stored in a s spring is given by:

E = ½Ke²

Where

E => is the energy

K => is the spring constant

e => is the elongation

From:

E = ½Ke²

Energy is directly proportional to the elongation. Thus,

E₁/e₁² = E₂/e₂²

With the above formula, we can obtain the final energy as follow:

Initial elongation (e₁) = 0.04 m

Initial energy (E₁) = E

Final elongation (e₂) = 0.08 m

Final energy (E₂) =?

E₁/e₁² = E₂/e₂²

E / 0.04² = E₂ / 0.08²

E / 0.0016 = E₂ / 0.0064

Cross multiply

0.0016 × E₂ = 0.0064E

Divide both side by 0.0016

E₂ = 0.0064E / 0.0016

E₂ = 4E

Therefore, the final energy is 4 times the initial energy i.e 4E

6 0
3 years ago
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