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Akimi4
17 hours ago
13

Peregrine falcons are known for their maneuvering ability. In a tight circular turn, a falcon can attain a centripetal accelerat

ion 1.5 times the free-fall acceleration. What is the radius of the turn if the falcon is flying at 17 m/s ?
Physics
1 answer:
Maru [1K]16 hours ago
8 0

Answer:

19.7 m

Explanation:

The centripetal acceleration experienced by the falcon can be expressed as

a=\frac{v^2}{r}

In this equation:

a represents acceleration

v denotes speed

r is the radius of the circular path

For this problem, we have:

a=1.5 g = 1.5(9.8)=14.7 m/s^2 as the acceleration

v = 17 m/s for the speed

By calculating r, we can determine the radius of the curve:

r=\frac{v^2}{a}=\frac{17^2}{14.7}=19.7 m

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Which model is used to describe the interaction of external forces that affect an organization's strategy and ability to compete
ValentinkaMS [1149]

Answer:

Competitive forces model

Explanation:

The Competitive forces model is a crucial instrument in strategic analysis aiming to assess an organization’s competitiveness. Commonly referred to as the "Five Force Model of Porter", this framework includes five key factors: the intensity of rivalry among existing competitors, the negotiating power of buyers, the threat posed by potential new entrants, the bargaining strength of suppliers, and the risk of substitute products or services.

These elements significantly influence an organization's competitive strategy and its likelihood of success.

5 0
4 days ago
A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to
Sav [1105]

Answer:

529.15 m/s

Explanation:

h = Highest point = 70000 m

g = Gravitational acceleration = 2 m/s²

m = Sulfur's mass

Since both potential and kinetic energies are conserved

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s

The velocity at which the liquid sulfur exited the volcano is 529.15 m/s

7 0
3 days ago
A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i
inna [987]

Answer:

The ratio of mass that is discarded is determined by this equation:

M - m = (3a/2)/(g²- (a²/2) - (ag/2))

Explanation:

The force acting on an object in motion is defined by the equation:

F = ma

Additionally, there is a gravitational force consistently acting downwards on the object, defined as g = 9.8 ms⁻²

For convenience, we will utilize a positive notation for downward acceleration and a negative notation for upward acceleration.

Case 1:

The hot air balloon has mass = M

Acceleration = a

Upward thrust from hot air = F = constant

Gravitational force acting downward = Mg

The net force on the balloon can be expressed as:

Ma = Gravitational force - Upward Force                              

Ma = Mg - F                      (since the balloon moves downward, that means Mg > F)

F = Mg - Ma

F = M (g-a)

M = F/(g-a)

Case 2:

After releasing the ballast, the new mass becomes m. The new upward acceleration is -a/2:

The net force is expressed as:

-m(a/2) = mg - F        (The balloon is moving upwards, hence F > mg)

F = mg + m(a/2)

F = m(g + (a/2))

m = F/(g + (a/2))

Determining the fraction of the mass initially dropped:

M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]

5 0
1 day ago
Consider a perfectly insulated cup (no
ValentinkaMS [1149]

Answer:

When ice is subjected to heat, it melts; however, the temperature remains constant at 0◦ C.

Explanation:

Solution

The heat supplied by the heater is solely utilized for the melting of the ice, thus maintaining the temperature at 0◦ C.

Once all the ice has liquefied, the temperature of the resulting water will start to rise over time.

Note: please see the attached document with solutions featuring diagrams related to this explanation

7 0
15 days ago
A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At
kicyunya [1025]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + \frac{1}{2}at^{2}

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + \frac{1}{2}(2.25 x 15.4^{2})

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

v^{2} = u^{2} + 2gs

0^{2} = 34.65^{2} + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying v^{2} = u^{2} + 2gs

v^{2} = 0^{2} + 2(9.81 x 328)

v^{2} = 0 + 6435.36

v = \sqrt{6435.36} = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

7 0
5 days ago
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