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victus00 [196]
4 years ago
15

If the spring constant is doubled, what value does the period have for a mass on a spring?

Physics
2 answers:
Damm [24]4 years ago
7 0
If spring constant is doubled, the mass on spring will be doubled as well. according to this formula, F=ke
k stands for spring constant and e stands for the length extended
Nat2105 [25]4 years ago
5 0

Answer:

T' = T / sqrt 2

Explanation:

According to the formula of time period of a loaded spring

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

where, m is the mass on the spring and k be the spring constant.

Here we observe that the time period is inversely proportional to the square root of the spring constant.

If the spring constant is doubled then the time period is

T' = 2\pi \sqrt{\frac{m}{2k}}     ....(2)

Dividing equation (2) by equation (1), we get

T' = T / sqrt 2

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Answer:

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There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co
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Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

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F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

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Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be
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consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s         (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

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