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frozen [14]
3 years ago
5

A train can travel at 300km/h. How long will it take to to travel 12.000km?

Physics
2 answers:
Leokris [45]3 years ago
8 0

Answer:

Train will take 156 sec to travel 12km.

Amanda [17]3 years ago
8 0
156 seconds or 2 minutes & 36 seconds
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Consider the following distribution of objects: a 2.00-kg object with its center of gravity at (0, 0) m, a 2.20-kg object at (0,
adelina 88 [10]

Answer:

body position 4 is (-1,133, -1.83)

Explanation:

The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by

               x_cm = 1 /M   ∑ x_{i} m_{i}

               y_cm = 1 /M   ∑ y_{i} mi

Where M is the total mass of the body, mi is the mass of each element

give us the mass and position of this masses

body 1

m1 = 2.00 ka

x1 = 0 me

y1 = 0 me

body 2

m2 = 2.20 kg

x2 = 0m

y2 = 5 m

body 3

m3 = 3.4 kg

x3 = 2.00 m

y3 = 0

body 4

m4 = 6 kg

    x4=?

   y4=?

mass center position

x_cm = 0

y_cm = 0

let's apply to the equations of the initial part

X axis

    M = 2.00 + 2.20 + 3.40

    M = 7.6 kg

    0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)

     x4 = -6.8 / 6

     x4 = -1,133 m

Axis y

    0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)

    y4 = -11/6

    y4 = -1.83 m

body position 4 is (-1,133, -1.83)

7 0
4 years ago
A 2150 kg satellite used in a cellular telephone network is in a circular orbit at a height of 780 km above the surface of the e
Tom [10]

Answer:

a)F=16741.9N

b)\frac{F}{W}=0.795

Explanation:

The gravitational force on the satellite is calculated with Newton's Gravitation Law:

F=\frac{GMm}{r^2}

Where M=5.97\times10^{24}kg is Earth's mass, m=2150kg is the satellite mass, r=R+h is the distance between their centers, where h=780000m is the height of the satellite (from Earth's surface) and R=6371000m is Earth's radius, and G=6.67\times10^{-11}Nm^2/kg^2 is the gravitational constant.

a) With these values we then have:

F=\frac{GMm}{r^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)(2150kg)}{(6371000m+780000m)^2}=16741.9N

b) And the fraction this force is of the satellite’s weight <em>W=mg</em> is:

\frac{F}{W}=\frac{GMm}{mgr^2}=\frac{GM}{gr^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)}{(9.8m/s^2)(6371000m+780000m)^2}=0.795

5 0
4 years ago
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