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2 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]2 years ago

5 0

57,600 cm/s ^2 ...........

denis-greek [22]2 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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A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficien

Liula [17]

Answer:

Yes it will move and a= 4.19m/s^2

Explanation:

In order for the box to move it needs to overcome the maximum static friction force

Max Static Friction = μFn(normal force)

plug in givens

Max Static friction = 31.9226

Since 36.6>31.9226, the box will move

Mass= Wieght/g which is 45.8/9.8= 4.67kg

Fnet = Fapp-Fk

= 36.6-16.9918

=19.6082

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Solve for a=4.19m/s^2

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3 years ago

A hand lifts a block vertically upward at constant velocity. The work done by gravity on the block ____ if the system consists o

adelina 88 [10]

Answer:

If the system consists of the block only, the work done by the gravity is negative.

If the system consists of the block and the earth the work done by the gravity is zero.

Explanation:

If the system consists of the block only, then the system experiences two external forces: one exerted by the hand that lifts the block vertically upward and other exerted by the earth (gravity), which is opposed to the movement of the system, so the work done by gravity is negative.

On the other hand, if the system consists of the block and the earth, then only exists a external force which is the exerted by the hand. So, the force exerted by gravity is zero.

6 0

3 years ago

A pressure of 400 Pa is applied to an area of 2.5 m2.What force applies this pressure?

Irina-Kira [14]

F = 400 Pa x 2.5 m2
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2 years ago

he adventurous robot M.A.N.D.I. is orbiting Saturn’s moon Dione. She wants to cause an impact with themoon to kick up some of th

GuDViN [60]

Answer:

v = 2.928 10³ m / s

Explanation:

For this exercise we use Newton's second law where the force is the gravitational pull force

F = ma

a = F / m

Acceleration is

a = dv / dt

a = dv / dr dr / dt

a = dv / dr v

v dv = a dr

We substitute

v dv = a dr

∫ v dv = 1 / m G m M ∫ 1 / r² dr

We integrate

½ v² = G M (-1 / r)

We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon

v² = 2G M (- 1 / R +2.73 10³+ 1 / R)

We calculate

v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61 - 10⁻³ /(5.61 + 2.73))

v² = 14.6828 10⁷ (0.1783 -0.1199)

v = √8.5748 10⁶

v = 2.928 10³ m / s

5 0

3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

10 months ago