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kap26 [50]
4 years ago
14

At what distance above Earth would a satellite have a period of 125 minutes

Physics
1 answer:
mash [69]4 years ago
5 0

Answer:

If a satellite orbiting the Earth has a period of 125 min, then it must be 1.90 x 106 m above the Earth's surface.

Explanation:

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2. The force between charges is 7 N. The distance between the charges is 4 x 10-6 m. If one of the charges is 2 x 10-8 C, what i
g100num [7]

Answer:

2) 6.22 × 10^-13

3) 4.56 × 10^-4

Explanation:

4 0
3 years ago
A toy rocket that weights 10 N blasts straight up from ground level with an initial kinetic energy of 40J. at the exact top of i
marishachu [46]

Answer:

14 m

Explanation:

Applying,

P.E = WH............ Equation 1

Where P.E = potential energy at the highest vertical height, W = weight of the toy rocket, H = Vertical height.

Note: At the highest vertical height, the kinetic energy is 0 J

Therefore,

Mechanical energy(M.E) = Potential Energy(P.E)

make H the subject of the equation

H = P.E/W......... Equation 2

From the question,

Given: W = 10 N,  P.E = 140 J

Substitute these values in equation 2

H = 140/10

H = 14 m

8 0
3 years ago
Why do you think there are so many more autotrophs than consumers on Earth?<br> Helppp
goldfiish [28.3K]

There are many Autotrophs than consumers because : Autotrophs are the bases of any food chain/web

<h3>Function of Autotrophs </h3>

Autotrophs capture the energy required for every food chain or web from which sustains other organisms in the food chain form sunlight and chemicals. Also there is always more biomass present in the lower levels of the trophic system as biomass decreases as we move up the food chain.

Since autotrophs are producers, for a healthy food chain there would be more autotrophs than consumers.

Hence we can conclude that There are many Autotrophs than consumers because : Autotrophs are the bases of any food chain/web

Learn more about Autotrophs : brainly.com/question/10253663

#SPJ1

3 0
2 years ago
Question # 40
jenyasd209 [6]

Answer:

Potential Energy = x = m g h

Kinetic energy = 1/2 m v^2

Assuming the mass fall from rest

1/2 m v^2 = m g h

v^2 = 2 g h

So the speed attained is independent of the mass

Also, x / v   does not have the units of mass

So the solution is none of the above.

7 0
3 years ago
One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicula
qwelly [4]

Answer:

Bnet=1.006*10^-6T

Explanation:

One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?

the magnetic field Bnet=\sqrt{b1^2+b2^2}

the magnetic field due this long wire is given by

B1=∨I1/(2\pi *R1)..............................1

B2=∨I2/(2\pi *R2)............................2

Bnet=\sqrt{(vI1/2*pi*R1)^2+(vI2/2*pi*R2)^2}.......................3

Bnet=v/2*pi\sqrt{(I1/R1)^2+(i2/R2)^2}

Bnet=4*pi*10^-7/(2\pi)\sqrt{(43/1.7)^2+(41/29.5)^2}

Bnet=0.0000002*(641.72)^.5

Bnet=1.006*10^-6T

8 0
4 years ago
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