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3 years ago

How many words can be made with five letters

Physics

2 answers:

dangina [55]3 years ago

6 0

The last word is gear

Firdavs [7]3 years ago

4 0

Answer:

agree eager eagre ragee

Explanation:

hope this help

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A 150 N boy rides a 60 N bicycle a total of 200 m at a constant speed. The frictional force against the forward motion of the bi

soldier1979 [14.2K]

Answer:

W = 7000 J

Explanation:

To solve this problem we use that the speed of the bicycle is constant, therefore its acceleration is zero

F -fr = 0

F = fr

where F is the force applied by the child

Work is defined by

W = F. x

W = F x cos θ

in this case the child's force is parallel to the movement, therefore the angle is zero and cos 0 = 1

let's calculate

W = 35 200

W = 7000 J

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3 years ago

Why are light intensity, carbon dioxide concentration and temperature liimiting factors

Harman [31]

Answer:

As carbon dioxide concentrations increase, so too does the rate of photosynthesis until a certain point where the graph levels off. At lower carbon dioxide concentrations carbon dioxide is the limiting factor because an increase in carbon dioxide causes an increase in photosynthesis.

Explanation:

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3 years ago

Why is weathering slow in cold, dry places?

Alexxandr [17]

Rate of weathering depends on temperature and moisture. Cold dry places have less water to weather things

Hope this helped!

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3 years ago

Based on the table above, _______ planets are made of the same things as _______ moons.

wolverine [178]

Answer:

Explanation:

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3 years ago

Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o

Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

$U=-\frac{GMm}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

$\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

and so, substituting:

$R=6370 km\\h_A = 5970 km\\h_B = 21200 km$

We find

$\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448$

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

$K=\frac{1}{2}\frac{GMm}{r}$

So, the ratio between the two kinetic energies is

$\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}$

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

$E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A, we have

$E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J$

For satellite B, we have

$E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J$

So, satellite B has the greater total energy (since the energy is negative).

(d) $-2.57\cdot 10^8 J$

The difference between the energy of the two satellites is:

$E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J$

4 0

3 years ago