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timama [110]
2 years ago
5

Two equal masses are attached to separate identical springs next to one another. One mass is pulled so its spring stretches 20 c

m and the other is pulled so its spring stretches only 10 cm. The masses are released simultaneously.Which mass reaches the equilibrium point first?
Physics
2 answers:
wlad13 [49]2 years ago
7 0

Answer:

Both reaches equilibrium simultaneously.

Explanation:

Given that,

First stretch spring= 20 cm

Second stretch spring = 10 cm

Two equal masses are suspended from identical spring.

Their forces constants are equal time period for oscillation

Using formula of time period

T=2\pi\sqrt{\dfrac{m}{k}}

Where, T = time period

m = mass

k = spring constant

Time period independent of amplitude.

Hence, Both reaches equilibrium simultaneously.

vazorg [7]2 years ago
7 0

Answer:

both reaches at the same time.

Explanation:

The time period of the mass is given by

T=2\pi \sqrt{\frac{m}{K}}

where, K be the spring constant and m be the mass attached.

As both the masses are same and springs are identical so K is same for both the springs.

Thus, both the masses will reach at the equilibrium position at the same time.

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A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte
Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

e = 4.9 cm

3 0
2 years ago
Fiber optic (FO) cables are based upon the concept of total internal reflection (TIR), which is achieved when the FO core and cl
kozerog [31]

Answer:

False

Explanation:

Though fiber active cable is based on the concept of internal reflection but it is achieved by refractive index which transmit data through fast traveling pulses of light. It has a layer of glass and insulating casing called “cladding,”and this is is wrapped around the central fiber thereby causing light to continuously bounce back from the walls of the Cable.

7 0
2 years ago
A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block
solniwko [45]
We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
4 0
2 years ago
How many times could Haley fly bewteen the two flowers in 1 minute (60 seconds)​
dmitriy555 [2]
30 seconds is the answer
7 0
2 years ago
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Two planes leave wichita at noon. one plane flies east 30 mi/h faster than the other plane, which is flying west. at what time w
scoundrel [369]
You have to take note of the individual directions of the plane. Since one is heading east, and the other is heading west, the planes are heading at opposite directions. So, it means that their distance between each other would be equal to 1,200 miles which accounts for the sum of their individual distances. The equation is as follows:

Total Distance = Distance of slower plane + Distance of faster plane
1,200 miles = st + (30+s)(t)
where
s is the speed of the slower plane and t is the time. Since both are not given, the final answer would just be in terms of s.
1,200 = t(s + 30 + s)
t = 1200/(30+2s)
t = 600/(15+s)
4 0
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