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REY [17]
3 years ago
14

A hummingbird flies in circles around a bush. if the velocity of the hummingbird is 720 cm/s and the radius of the circular path

is 9.0 cm, what is the centripetal acceleration of the hummingbird?
Physics
2 answers:
Rashid [163]3 years ago
5 0
57,600 cm/s ^2 ...........
denis-greek [22]3 years ago
4 0

Answer:

57600cm/s²

Explanation:

centripetal accleration best describes the motion of the humming bird. it is the motion inwards towards the centre of a circle.

centripetal acceleration (a_{c}) = \frac{v^{2} }{r}

where v= velocity = 720cm/s

r=9cm

r= radius

a_{c}= 720² ÷ 9

a_{c}= \frac{720 * 720}{9}= 57600cm/s²

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Answer:

An investigation is made to determine the performance of simple thin airfoils in the slightly supersonic flow region with the aid of the nonlinear transonic theory first developed by von Kármán[1]. Expressions for the pressure coefficient across an oblique shock and a Prandtl-Meyer expansion are developed in terms of a transonic similarity parameter. Aerodynamic coefficients are calculated in similarity form for the flat plate and asymmetric wedge airfoils, and curves are plotted. Sample curves for a flat plate and a specific asymmetric wedge are plotted on the usual coordinate grid of Cl, Cd,andCmc/4versus angle of attack and Cl versus Mach Number to illustrate the apparent features of nonlinear flow.

Explanation:

5 0
3 years ago
If two 100 ohms resistors are placed in series, their total resistance is what?
Sindrei [870]

Answer:

add the Resistance

2(100)= 200

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3 0
3 years ago
How much 8 grams will weigh after a physical change
Ber [7]
I would look this one up on Google
5 0
2 years ago
A 20-kg child running at 1.4 m/s jumps onto a playground merry-go-round that has mass 180 kg and radius 1.6m. She is moving tang
Dominik [7]

Answer:

ωf = 0.16 rad/s

Explanation:

Moment of inertia of the child = mr² = 20(1.6²) = 51.2 kg•m²

Moment of Inertia of the MGR = ½mr² = ½(180)1.6² = 230.4 kg•m²

(ASSUMING it is a uniform disk)

Initial angular momentum of the child = Iω = I(v/r) = 51.2(1.4/1.6) = 44.8 kg•m²/s

Conservation of angular momentum

44.8 = (51.2 + 230.4)ωf

ωf = 0.15909090...

4 0
2 years ago
A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1
kicyunya [14]

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

\mu = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s

Frequency is given by

f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz

Angular frequency is given by

\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s

Maximum velocity of a particle is given by

v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s

The maximum velocity of a particle on the string is 8.89288275 m/s

5 0
3 years ago
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