All categories

2 years ago

A hummingbird can a flutter its wings 4800 times per minute what is the frequency of wing flutters per second

Physics

2 answers:

Scilla [17]2 years ago

7 0

the answer i got is 80 i hope this helpss!!!!

Arlecino [84]2 years ago

7 0

Answer:

The frequency of wing flutters per second is 80.

Explanation:

Frequency is defined as the number of waves that passes a given point per second. It is expressed in Hertz that second inverse.

$1 Hertz = 1 s^{-1}$

Number of times wings fluttered by hummingbird = $4800 min^{-1}$

1 minute = 60 seconds

Frequency wing flutters per second = F

$F = 4800 min^{-1}=\frac{4800}{60 s}=80 s^{-1}$

The frequency of wing flutters per second is 80.

You might be interested in

The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio

enot [183]

Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean ( $\mu$ ) plus or minus one standard deviation ( $\sigma$ ), $P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%$

at the range of mean plus or minus two standard deviation, $P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%$

at the range of mean plus or minus three standard deviation, $P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%$

So, note that they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) ( $P(X \leq \mu+2\sigma)$ )

So we know that the 95.5% is between $\mu - 2\sigma = 0.3 -2*0.1 = 0.28$ and $\mu + 2\sigma = 0.3 +2*0.1 = 0.32$ , hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than 0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

So $P(X \leq \mu+2\sigma) \approx 2.25%$

8 0

2 years ago

A player kicks a football into the air. It slows to a stop at its highest point in the air before falling to the ground. Which s

vivado [14]

Answer:

The ball slows down in the air due to an unbalanced force

Explanation:

When player kicks the ball, there are mainly two foces acting on this object: the force made by the player and the opposite force of gravity (which acts with a direction always to the centre of the Earth)

The force applied by the player will be decreasing, while the force of gravity is always constant, this will make that both forces will unbalance, making the football´s speed slow down

8 0

2 years ago

Read 2 more answers

Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous

Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is Av with A the cross section area and v the velocity,

so,

$A_1V_1=A_2V_2$

$A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}$

the diameter decreases 86% so

$d_2 = 0.86d_1$

$v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}$

Thus, speed in smaller section is 48.6 m/s

3 0

2 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

skelet666 [1.2K]

Answer:

So the acceleration of the child will be $8.05m/sec^2$

Explanation:

We have given angular speed of the child $\omega =1.25rad/sec$

Radius r = 4.65 m

Angular acceleration $\alpha =0.745rad/sec^2$

We know that linear velocity is given by $v=\omega r=1.25\times 4.65=5.815m/sec$

We know that radial acceleration is given by $a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2$

Tangential acceleration is given by

$a_t=\alpha r=0.745\times 4.65=3.464m/sec^$

So total acceleration will be $a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2$

7 0

2 years ago

A dog of mass 10 kg sits on a skateboard of mass 2 kg that is initially traveling south at 2 m/s. The dog jumps off with a veloc

Tasya [4]

Answer:

17 m/s south

Explanation:

$m_1$ = Mass of dog = 10 kg

$m_2$ = Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

$u_1$ = Velocity of dog = 1 m/s

$u_2$ = Velocity of skateboard

In this system the linear momentum is conserved

$(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s$

The velocity of the skateboard will be 17 m/s south as the north is taken as positive

3 0

2 years ago