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vovikov84 [41]
3 years ago
7

Two identical 0.200kg mass are pressed against opposite ends of a light spring of force constant 1.75N/cm compressing the spring

by 37 cm from it’s normal length . Find the speed of each mass when it has moved free of the spring on a frictionless horizontal table
Physics
1 answer:
arlik [135]3 years ago
4 0

This type of a problem can be solved by considering energy transformations. Initially, the spring is compressed, thus having stored something called an elastic potential energy. This energy is proportional to the square of the spring displacement d from its normal (neutral position) and the spring constant k:

E_p=\frac{1}{2}kd^2= \frac{1}{2}175\frac{N}{m}\cdot 0.37^2m^2=11.98J

So, this spring is storing almost 12 Joules of potential energy. This energy is ready to be transformed into the kinetic energy when the masses are released. There are two 0.2kg masses that will be moving away from each other, their total kinetic energy after the release equaling the elastic energy prior to the release (no losses, since there is no friction to be reckoned with).

The kinetic energy of a mass m moving with a velocity v is given by:

E_k = \frac{1}{2}mv^2

And we know that the energies are conserved, so the two kinetic energies will equal the elastic potential one:

E_p = 2E_k=mv^2

From this we can determine the speed of the mass:

E_p =mv^2\implies v=\pm \sqrt{\frac{E_p}{m}}=\pm\sqrt{\frac{11.98J}{0.2kg}}=\pm 7.74\frac{m}{s}

The speed will be 7.74m/s in in one direction (+), and same magnitude in the opposite direction (-).

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6 0
3 years ago
Read 2 more answers
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light
motikmotik

Answer:

n_cladding = 1.4764

Explanation:

We are told that θ_max = 5 °

Thus;

θ_max + θ_c = 90°

θ_c = 90° - θ_max

θ_c = 90° - 5°

θ_c = 85°

Now, critical angle is given by;

θ_c = sin^(-1) (n_cladding/n_core)

sin θ_c = (n_cladding/n_core)

n_cladding = (n_core) × sin θ_c

Plugging in the relevant values, we have;

n_cladding = 1.482 × sin 85

n_cladding = 1.4764

7 0
3 years ago
Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.05 mm high. Assume it starts from rest
mestny [16]

Height is 7.05 m and not 7.05 mm

Answer:

9.603 m/s

Explanation:

We are dealing with rotation, so velocity of centre of mass is given by;

v_cm = Rω

Since we are working with a solid cylinder, moment of inertia of the cylinder is; I = ½mR²

Since it is rolled from the top to the bottom, at the top it will have potential energy(mgh) while at the bottom it will have kinetic energy (rotational plus translational kinetic energy).

Using conservation of energy, we have:

P.E = K.E_t + K.E_r

Formula for rotational and kinetic energy here are;

K.E_t = ½mv²

K.E_r = ½Iω²

mgh = ½mv² + ½Iω²

Since we want to find translational speed(v), let's get rid of ω.

Earlier, we saw that v_cm = Rω

Thus; ω = v/R

Also, we know that I = ½mR².

Thus;

mgh = ½mv² + ½(½mR²)(v/R)²

This gives;

mgh = ½mv² + ¼mv²

Divide through by m to get;

gh = v²(½ + ¼)

gh = ¾v²

Making v the subject gives;

v = √(4gh/3)

v = √((4 × 9.81 × 7.05)/3)

v = 9.603 m/s

6 0
2 years ago
Why is the picture above not an accurate representation of charges and their field lines?
pochemuha
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7 0
2 years ago
Read 2 more answers
A bowling ball is dropped off the top of the Eiffel Tower. If the Eiffel Tower is 300 meters
patriot [66]

Answer:

7.82 s

Explanation:

Given:

Δy = 300 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(300 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 7.82 s

5 0
3 years ago
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