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Firlakuza [10]
3 years ago
8

A 2.5 kg tribble is placed in a bucket and whirled in a 1.4 m radius vertical circle at a constant tangential speed. If the forc

e on the tribble from the bucket does not exceed 10 times its weight (tribbles are a hardy species), the maximum tangential speed is
Physics
1 answer:
Over [174]3 years ago
8 0

Given that,

Mass of a tribble, m = 2.5 kg

Radius, r = 1.4 m

The force on the tribble from the bucket does not exceed 10 times its weight.

To find,

The maximum tangential speed.

Solution,

The force acting on the tribble is equal to the centripetal force.

F = 10mg

The formula for the centripetal force is given by :

F=\dfrac{mv^2}{r}

v is maximum tangential speed

v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{mgr}{m}} \\\\v=\sqrt{{10gr}} \\\\v=\sqrt{10\times 9.8\times 1.4} \\\\v=11.7\ m/s

So, the maximum tangential speed is 11.7 m/s.

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The Doppler effect causes sounds moving toward an observer to sound _____.
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Answer: softer

The sound waves become weaker, so to an observer this wave is lighter sound.

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3 years ago
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X<br> x<br> 2.5 N<br> 1.7 N<br> 2.5 N
Olin [163]

Answer:

2.5*1.7/2.5

Explanation:

6 0
3 years ago
A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ
Svetradugi [14.3K]

Answer:

d = 4180.3m

wavelengt of sound is 0.251m

Explanation:

Given that

frequency of the sound is 5920 Hz

v=1485m/s

t=5.63s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}

d=\frac{1485*5.63}{2}\\d= 4180.3m

wavelengt of sound is \lambda = v/f

= (1485)/(5920)

= 0.251 m

7 0
3 years ago
A boy whirls a stone in a horizontal circle of radius 1.4 m and at height 1.5 m above ground level. The string breaks, and the s
balandron [24]

Answer:233.23 m/s^2

Explanation:

Given

radius of circle=1.4 m

Height of stone above ground=1.5 m

Horizontal distance(R)=10 m

It is given at the time of break stone flies horizontally thus stone to cover a height of 1.5 m in time t before reaching ground

1.5=0+\frac{gt^2}{2}

t=0.55 s

Initial horizontal velocity at the time of break is given by u

R=u\times t

10=u\times 0.55

u=18.07 m/s

Therefore magnitude of centripetal acceleration is given by

a_c=\frac{u^2}{r}=\frac{18.07^2}{1.4}=233.23 m/s^2

6 0
3 years ago
At what speed does a 1400 compact car have the same kinetic energy as a 20000 truck going 25 ?
kumpel [21]
K=1400*V^2/2
K=20000*25^2/2. => 1400*V^2/2=20000*25^2/2 <=> 1400*V^2=20000*25^2
14*V^2=200*225
v^2=100*225/7
v=250/7^(1/2)

Answer: 250*7^(1/2)/7
4 0
3 years ago
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