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olganol [36]
3 years ago
9

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons

tant speed, it would pass 24000 km from the center of Earth. However, it will be attracted to Earth and might hit our planet. What is the minimum speed the asteroid should have so it will just graze the surface of the Earth?
Physics
1 answer:
Vlada [557]3 years ago
5 0

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

K.E\geq P.E

or

\dfrac{1}{2}mv^2 \geq  G\dfrac{Mm}{R}

where m is the mass of the asteroid, R= 24,000,000\:m is its distance form earth's center, M = 5.9*10^{24}kg is the mass of the earth, and G = 6.7*10^{-11}m^3/kg\: s^2 is the gravitational constant.

Solving for v we get:

v \geq \sqrt{\dfrac{2GM}{R} }

putting in numerical values gives

v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }

\boxed{v\geq 5739.5m/s}

in kilometers this is

v\geq5.7395m/s.

Hence, the minimum speed required is 5.7395km/s.

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A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ
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Explanation:

The given data is as follows.

                    mass = 0.20 kg

              displacement = 2.6 cm

              Kinetic energy = 1.4 J

       Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

         Total energy = Kinetic energy + spring potential energy

                           = 1.4 J + 2.2 J

                            = 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So,      K.E = Total energy

                = 3.6 J

Also, we know that

                  K.E = \frac{1}{2}mv^{2}_{m}

or,                   v = \sqrt{\frac{2K.E}{m}}

                        = \sqrt{2 \times 3.6 J}{0.2 kg}

                        = \sqrt{36}

                        = 6 m/s

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A parallel-plate air capacitor of area A = 15.0 cm2 and plate separation d = 3.00 mm is charged by a battery to a voltage 58.0 V
sladkih [1.3K]

Answer:

The additional charge that will flow from the battery onto the positive plate is 0.924 nC

Explanation:

Given;

Area of the capacitor, A = 15.0 cm² = 15 x 10⁻⁴ m²

Separation distance, d = 3.00 mm = 3 x 10⁻³ m²

voltage of the capacitor, V = 58.0 V

dielectric constant, k = 4.60

Initial Capacitance of the capacitor before the addition of dielectric material:

C = \frac{\epsilon _oA}{d} = \frac{8.85*10^{-12}*15*10^{-4}}{3*10^{-3}} = 4.425 *10^{-12} F

Initial charge across the parallel plates:

Q₁ = CV

Q = 4.425 x 10⁻¹² x 58 = 2.567 x 10⁻¹⁰ C

Capacitance of the capacitor after the addition of dielectric material:

Cequ. = Ck

Cequ. = 4.425 x 10⁻¹²F x 4.6 = 20.355 x 10⁻¹² F

Final charge across the parallel plates:

Q₂ = Cequ. x V

Q₂ =  20.355 x 10⁻¹² x 58 = 11.806 x 10⁻¹⁰ C

Additional charge = Q₂ - Q₁

                              = 11.806 x 10⁻¹⁰ C - 2.567 x 10⁻¹⁰ C

                              = 9.239 x 10⁻¹⁰ C

                              = 0.924 nC

Therefore, the additional charge that will flow from the battery onto the positive plate is 0.924 nC

3 0
3 years ago
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