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Vera_Pavlovna [14]
3 years ago
5

A bird has a mass of 0.8 kg and flies at a speed of 11.2 m/s. How much kinetic energy does the bird have?

Physics
2 answers:
riadik2000 [5.3K]3 years ago
7 0
The kinetic energy of an object is directly proportional to its mass and the square of its velocity

KE = 1/2 (mv²)

KE = Kinetic Energy
m = mass in kg
v = velocity in m/s

Given:

m = .8 kg
v =  11.2 m/s

Substitute:

KE = 1/2 (.8)(11.2²)
KE = 50.18 J
scZoUnD [109]3 years ago
5 0

Answer:

50.2 like the other dude said

Explanation:

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Using a flowchart, discuss how a tsunami develops?​
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A tsunami is a sequence of particularly long water waves that can spread over very great distances and, as such, cause water to move.

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4 0
3 years ago
A reconnaissance plane flies 556 km away from
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We use the formula, to calculate the average speed of the round trip,

v_{av} =\frac{d_{total}}{t_{total}}

Here, d_{total}, is total distance covered by plane in total time, t_{total}.

For the round trip,

d_{total=556\ km+556\ km=1112 \times 10^3\ m

t_{total}=\frac{556 \times 10^3 m}{832 m/s} + \frac{556 \times 10^3 m}{1248 m/s}.

Thus,

v_{av} =\frac{1112 \times 10^3\ m}{\frac{556 \times 10^3 m}{832 m/s} + \frac{556 \times 10^3 m}{1248 m/s}} =\frac{1112 \times 10^3\ m}{668.26\ s+445.51\s} \\\\v_{av}=998.41\ m/s.


5 0
4 years ago
Astronomers discover an exoplanet (a planet of a star other than the Sun) that has an orbital period of 3.87 Earth years in its
dolphi86 [110]

Answer: 4.487(10)^{11}m

Explanation:

This problem can be solved using the Third Kepler’s Law of Planetary motion:

<em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.  </em>

<em />

This law states a relation between the orbital period T of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size a of its orbit:

T^{2}=\frac{4\pi^{2}}{GM}a^{3} (1)  

Where:

T=3.87Earth-years=122044320s is the period of the orbit of the exoplanet (considering 1Earth-year=365days)

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}  

M=3.59(10)^{30}kg is the mass of the star

a is orbital radius of the orbit the exoplanet describes around its star.

Now, if we want to find the radius, we have to rewrite (1) as:

a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}} (2)  

a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}} (3)  

Finally:

a=4.487(10)^{11}m This is the radius of the exoplanet's orbit

3 0
3 years ago
Using .125 mass kgwhat is the velocity​
agasfer [191]

Answer:

The answer is 5 meters per second.

6 0
3 years ago
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