3 years ago

Function do lappet which faced vultures serve in a temperate grassland

Physics

2 answers:

vfiekz [6]3 years ago

7 0

Um this doesn't make since to me since you did not clearly state your awnser

Blababa [14]3 years ago

3 0

Answer:

free

Explanation:

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A 0.155 kg arrow is shot upward

solniwko [45]

Answer:

2.43J

Explanation:

Given parameters:

Mass of the arrow = 0.155kg

Velocity = 31.4m /s

Unknown:

Kinetic energy when it leaves the bow = ?

Solution:

The kinetic energy of a body is the energy in motion of the body;

it can be derived using the expression below:

K.E = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

Solve for K.E;

K.E = $\frac{1}{2}$ x 0.155 x 31.4 = 2.43J

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3 years ago

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The path the moon's umbra traces across earth's surface is the path of totality.What would you see if you were standing in the p

ivann1987 [24]

You would see the darkest part of the moon

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3 years ago

If a ball is thrown horizontally with a speed of 65 mph, how far will it fall while traveling 90 ft of horizontal distance?

kow [346]

Answer:

install socrati it give you all answers

Explanation:

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3 years ago

Newton’s second law states that the acceleration of an object is dependent on 1 variable which is the net force

Darya [45]

I think it’s false. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and then mass of the object.

7 0

3 years ago

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth = 5.98 × 10²⁴ kg

Radius of Earth = $6.38\times 10^6\ m$

The gravitational force will balance the centripetal force

$\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}$

$T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}$

From Kepler's law we have relation

$T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m$

Distance from the center of the Earth would be

$42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}$

8 0

3 years ago