Ask question

Ask question

All categories

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?

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

You might be interested in

In 1935, a French destroyer, La Terrible, attained one of the fastest speeds for any standard warship. Suppose it took 3.0 min a

REY [17]

Answer:

Explanation:

From newton's equation of motion of uniform acceleration

v = u + at

where v is final velocity , u is initial velocity , a is acceleration and time is t .

putting the values

v = 0 + .5 x 3 x 60 ( time in second = 3 x 60 s )

= 90 m /s

So , final velocity is 90 m /s .

7 0

3 years ago

The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of [L]3/[T]. The flow rate of li

melamori03 [73]

Answer: n=4

Explanation:

We have the following expression for the volume flow rate $Q$ of a hypodermic needle:

$Q=\frac{\pi R^{n}(P_{2}-P_{1})}{8\eta L}$ (1)

Where the dimensions of each one is:

Volume flow rate $Q=\frac{L^{3}}{T}$

Radius of the needle $R=L$

Length of the needle $L=L$

Pressures at opposite ends of the needle $P_{2}$ and $P_{1}=\frac{M}{LT^{2}}$

Viscosity of the liquid $\eta=\frac{M}{LT}$

We need to find the value of $n$ whicha has no dimensions, and in order to do this, we have to rewritte (1) with its dimensions:

$\frac{L^{3}}{T}=\frac{\pi L^{n}(\frac{M}{LT^{2}})}{8(\frac{M}{LT}) L}$ (2)

We need the right side of the equation to be equal to the left side of the equation (in dimensions):

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n}}{LT}$ (3)

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{n-1}}{T}$ (4)

As we can see $n$ must be 4 if we want the exponent to be 3:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{4-1}}{T}$ (5)

Finally:

$\frac{L^{3}}{T}=\frac{\pi}{8} \frac{ L^{3}}{T}$ (6)

8 0

3 years ago

Do you think there might be a point in space between Earth and the Moon where the gravity of each would pull on an object equall

Firlakuza [10]

Answer: Yes.

Explanation:

Assuming Earth and Moon are isolated is space, it is possible to have a point where Earth and Moon will pull at an object with equal force.

That point will be closer to the Moon than the Earth because Moon's gravitational field strength is weaker than Earth's gravitational field strength.

7 0

3 years ago

Seatbelts are necessary to prevent your body from flying out of a vehicle during a crash. why does your body not stop with the c

Svetlanka [38]

Because you are separate from the car.

6 0

3 years ago

The position of a 60 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.4

mrs_skeptik [129]

The position of an oscillating mass is given by:

$x(t)=A cos (\omega t)$

where A is the amplitude of the oscillation, $\omega$ the angular frequency and t the time.

The velocity of the oscillating mass can be found by calculating the derivative of the position:

$v(t)=x'(t)=-\omega A sin (\omega t)$

In this problem, A=2.0 cm and $\omega=10 rad/s$ , so if we substitute these data and t=0.4 s we can find the velocity at t=0.4 s:

$v(t)=-(10 rad/s)(2.0 cm) sin ((10 rad/s)(0.4 s))=-13.07 cm/s=-0.13 m/s$

3 0

3 years ago