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N76 [4]
3 years ago
11

An unladen swallow that weighs 0.03 kg flies straight northeast a distance of 125 km in 4.0 hours. With the x x direction due ea

st and the y y direction due north, what is the average momentum of the bird (in unit vector notation)?
Physics
1 answer:
balandron [24]3 years ago
8 0

Answer:

The average momentum of the bird is 0.26 kgm/s

Explanation:

The formula to be used here is that of momentum which is

momentum (in kgm/s) = mass (in kg) × velocity (in m/s)

The velocity of the bird is

velocity (in m/s) = distance (in meter) ÷ time (in seconds)

distance in meters = 125km × 1000 = 125,000 m

time in seconds = 4 hrs × 60 × 60 = 14,400 secs

velocity = 125000/14400

velocity = 8.68 m/s

momentum (p) = 0.03 × 8.68

p = 0.26 kgm/s

The average momentum of the bird is 0.26 kgm/s

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4 years ago
A camper stands in a valley between two parallel cliff walls. He claps his hands and notices that the echo from the nearby wall
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5,658 ft is your answer I believe
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2 years ago
Determine the total moment of inertia of a merry-go round with 5 children sitting on it. Of the five children, four are seated a
BaLLatris [955]

Answer:

Explanation:

Given that,

We have five children.

Each of mass m =30kg

They sit on a merry go round

Mass of Merry go round M= 150kg

Radius of Merry go round is r =2m

Four children sit at the edge of the merry go round but one child sit at the centre.

The four child that sit at the edge are 2m from the centre of the merry go round but the one at the centre is 0m from the centre

Moment of inertia?

Moment of inertia is given as

I=Σmi•ri²

For the question, the moment of inertia is the combination of inertial of child and the merry go round

I= I(merry go round) + I(four child)+ I(last child)

The merry go round is assumed to be a solid cylinder, so it is going to have the moment of inertia of solid cylinder

Then,

I(merry go round ) =½ Mr²

Also, Four of the child has the same moment of inertia, they are 2m form the centre of the merry go round why the last child has no moment of inertia

I= I(merry go round) + I(four child) +I(last child )

I= ½Mr² + 4mr² + mr'²

I = ½ × 150 ×2² + 4×30×2² + 30×0²

I = 300 +480+0

I = 780 kgm²

7 0
3 years ago
A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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Answer:

Person B has four times the power output of person A.

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