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Ivanshal [37]
3 years ago
12

A mass M is attached to a spring with spring constant k. When this system is set in motion with amplitude A, it has a period T.

What is the period if the mass is doubled to 2M
Physics
1 answer:
labwork [276]3 years ago
7 0

Answer:

T' = o.707T

Explanation:

T = 1/2\pi \sqrt{k/m} \\\\m'=2m then\\T' = 1/2\pi \sqrt{k/m'}\\    = 1/2\pi \sqrt{k/2m}\\T' = \frac{(1/2\pi \sqrt{k/m}\\)}{1.4} T' = \frac{T}{1.4}\\T' = 0.707T

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Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 2.4
NeX [460]

When person is observing destructive interference at 0.20 m distance from the equidistant position then we can say that path difference must be equal to half of the wavelength

now we will have

\frac{\lambda}{2} = \frac{yd}{L}

now we know that

y = 0.20 m

d = 2.4 m

L = 10 m

now here we have

\frac{\lambda}{2} = \frac{0.20\times 2.4}{10}

\lambda = 0.096 m

now frequency of wave is given as

f = \frac{v}{\lambda}

f = \frac{343}{0.096} = 3573 Hz

5 0
3 years ago
Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to v
butalik [34]

Answer:

u=34cm

Explanation:

From the question we are told that:

Far point is V=34 cm

Near point is u=17 cm

Therefore

Focal Length

f=-34cm

Generally the equation for the Lens is mathematically given by

\frac{1}{u}=\frac{1}{f}-\frac{1}{v}

\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}

u=34cm

5 0
3 years ago
10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth’s surface. What is its time perio
Afina-wow [57]

Answer:

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)

V2R+h=g(R2(R+h)2)

V=√g(R2R+h)

V= sqrt(9.8 × (6371000)^2/(6371000+360000)

V= sqrt(9.8× (4.059×10^13/6731000)

V=sqrt(65648789.18)

V= 8102.39m/s

Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)

T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)

T=sqrt(3.40×10^21)/ (3.99×10^14)

T= sqrt(0.862×10^7)

T= 2935.98seconds

4 0
3 years ago
Helppppppppppppppp.......
kow [346]

Answer:

9013 m/s

Explanation:

hope it helped!!!

8 0
3 years ago
A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con
erastova [34]

Let the key is free falling, therefore from equation of motion

h = ut +\frac{1}{2}gt^2..

Take initial velocity, u=0, so

h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2.

h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

d= v \times t

From above substituting t,

d = v \times \sqrt{\frac{2h}{g} }.

Now substituting all the given values and g = 9.8 m/s^2, we get

d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m.

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

7 0
3 years ago
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