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3 years ago

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What is Newton's second law of motion

1 answer:

Alexandra [31]3 years ago

8 0

Newton’s second law is that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

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Two satellites are in circular orbits around the earth. the orbit for satellite a is at a height of 540. km above the earth's su

kotegsom [21]

<span>Answer: Va = 7,625 m/s Vb = 7,404 m/s Given: A = 486,000 m B = 901,000 m G = 6.67428E-11 m^3/kg-s^2 M = 5.9736E+24 kg r = 6,371,000 m Recall that you need the actual orbital distance from the *center* of the Earth, giving radius plus altitude: rA = 6,857,000 m rB = 7,272,000 m Equation: V = SQRT { GM / r } Solve for A Va = SQRT { [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (6,857,000 m) } Va = SQRT { [ 3.9869 m^3/s^2 ] / (6,857,000 m) } Va = SQRT { 58,144,202 m^2/s^2 } Va = 7,625 m/s Solve for B Vb = SQRT { [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (7,272,000 m) } Vb = SQRT { [ 3.9869 m^3/s^2 ] / (7,272,000 m) } Vb = SQRT { 54,826,016 m^2/s^2 } Vb = 7,404 m/s</span>

7 0

3 years ago

What determines an object’s velocity?

joja [24]

Answer:

I think it's 3) speed and direction

8 0

4 years ago

If it is known that a non-zero net force is acting on an object, then which of the

Anna [14]

Answer:

I think it is other B,C I think not fur sure though

7 0

3 years ago

Two metal plates 15mm apart have a potential difference of 750v between them. The force on a small charged sphere placed between

Romashka [77]

Answer:

50,000 V/m

Explanation:

The electric field between two charged metal plates is uniform.

The relationship between potential difference and electric field strength for a uniform field is given by the equation

$\Delta V=Ed$

where

$\Delta V$ is the potential difference

E is the magnitude of the electric field

d is the distance between the plates

In this problem, we have:

$\Delta V=750 V$ is the potential difference between the plates

d = 15 mm = 0.015 m is the distance between the plates

Therefore, rearranging the equation we find the strength of the electric field:

$E=\frac{\Delta V}{d}=\frac{750}{0.015}=50,000 V/m$

6 0

4 years ago

An infinite slab of charge of thickness 2z0 lies in the xy-plane between z=−z0 and z=+z0. The volume charge density rho(C/m3) is

Oksi-84 [34.3K]

Answer:

please read the answer below

Explanation:

To find the electric field you can consider the Gaussian law for a cylindrical surface inside the slab.

$\int E dA=EA_{G}=\frac{Q_{int}}{\epsilon_o}$

$Q_{int}=\rho V_{G}$

where Qint is the charge inside the Gaussian surface, AG is the area of the surface and rho is the charge density of the slab.

By using the formula for the volume of a cylinder you obtain:

$V_{G}=\pi r^2h$

where h is the height. If you assume that the slab is in the interval (-zo<z<z0) you can write VG:

$V_{G}=\pi r^2 z$

Finally, by replacing in the expression for E you get:

$E=\frac{Q_{int}}{\epsilon_o A_G}=\frac{Q_{int}}{\epsilon_o \pi r^2}\frac{z}{z}=\frac{\rho z}{\epsilon_o}$

$E=\rho z/\epsilon_o$

hence, for z>0 you obtain E=pz/eo > 0

for z<0 -> E=pz/eo < 0

7 0

3 years ago