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kotykmax [81]
3 years ago
10

A meteoroid passes through a position in space where its speed is very small relative to Earth's and it is at a perpendicular di

stance of 19 Earth radii above Earth's surface. The meteoroid is moving in such a way that Earth captures it.
What is the speed of the meteoroid when it is one Earth radius above the ground
Physics
1 answer:
Oliga [24]3 years ago
3 0

Answer:

v=7506.4m/s

Explanation:

If we call 1 the position in space where the meteoroid of mass m speed is very small relative to Earth's (whose mass is M_E=5.98\times10^{24}kg and radius is R_E=6371000m) and it is at a perpendicular distance of h_1=19R_E above Earth's surface, and 2 the position when it is h_2=R_E above the ground, then, since in this case mechanical energy is conserved, we can write:

E_1=E_2

which means:

K_1+U_1=K_2+U_2

where K is the kinetic energy and U the gravitational potential energy. Since K_1=0J we can write:

\frac{-GM_Em}{r_1}=\frac{mv_2^2}{2}+\frac{-GM_Em}{r_2}

which means:

v_2=\sqrt{2GM_E(\frac{1}{r_2}-\frac{1}{r_1})}

And since r=R_E+h (the distance of an object to the center of the Earth is Earth's radius plus the height of the object), we have:

v_2=\sqrt{2GM_E(\frac{1}{R_E+h_2}-\frac{1}{R_E+h_1})}=\sqrt{2GM_E(\frac{1}{2R_E}-\frac{1}{20R_E})}=\sqrt{\frac{GM_E}{R_E}(1-\frac{1}{10})}=\sqrt{\frac{0.9GM_E}{R_E}}

This for our values is:

v_2=\sqrt{\frac{0.9GM_E}{R_E}}=\sqrt{\frac{0.9(6.67\times10^{-11}Nm^2/kg^2)(5.98\times10^{24}kg)}{6371000m}}=7506.4m/s

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Write about any 8 areas of field where physics is applied. illustrate with examples.
Ulleksa [173]

1. In construction. When constructing roads, building

2. In Engineering, when engineering we use physics such as calculations.

3. At home

4. In industries, industries use chemicals and so on.

5. In tailoring, we use the knowledge of physics like when sowing.

6. In carpentry, we use calculations.

7. In Hospitals, we use physics we balancing

8. In Sites, we use energy.

4 0
3 years ago
A Chinook salmon can swim underwater at 3.75 m/s, and it can also jump vertically upward, leaving the water with a speed of 6.20
Olin [163]

Answer:

P = 2161 N

Explanation:

For this exercise, let's start with Newton's second law

                P - W = m a

                P = ma + W

Where P is the fin force, W the weight,

Let's look for the vertical acceleration of the fish, this goes from a vertical speed of zero to a speed of 6.20 m / s when it has traveled half its length 1.50

                  y = 0.75 m

                  v² = v₀ + 2 a y

                  a = v² / 2 y

                  a = 6.20²/2 0.75

                  a = 25.63 m / s

We substitute in Newton's second law

                  P = m (g + a)

                 

Let's calculate

                 P = 61.0 (0 9.8 + 25.63)

                P = 2161 N

5 0
3 years ago
Is thin humans more denser than fats??
yuradex [85]
Thin because muscle is denser than fat
5 0
4 years ago
A toy train rolls around a horizontal 1.0-m-diameter track. the coefficient of rolling friction is 0.10. how long does it take t
Musya8 [376]
Deceleration due to rolling resistance (α) = coefficient*g/R = 0.1*9.81/0.5 = 1.962 rad/sec^2

Equations of motion:
ωf=ωo+αt
But ωf = 0 rad/sec; ωo= 2*pi*RPM/60 = 4.19 rad/sec

Therefore,
0=4.19-1.962t => 1.962t = 4.19 => t = 4.19/1.962 = 2.13 seconds
6 0
3 years ago
Read 2 more answers
1. Which wave refracts?
Digiron [165]

Answer:

I want to say your answer is A) an ocean wave approaching the shore at an angle. But not 100% sure. I'm so sorry if it's wrong I really tried to figure it out.

Explanation:

Refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refaction , or the bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves. Thus, if water waves are passing from deep water into shallow water, they will slow down. ( This is what I read on google maybe it'll help you out).

8 0
3 years ago
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