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Yuliya22 [10]
3 years ago
11

Change 1m2 in to cm2, mm2 and km2​

Physics
1 answer:
tresset_1 [31]3 years ago
7 0

Answer:

1m² = 10000cm²

1m² = 1000000mm²

1m² = 1 × 10^-6

Explanation:

When coverting from square meter to square centimetre, multiply the area value by 10.

When coverting from square meter to millimetre, multiply the area value by 10⁶ ( 1 million )

When converting from meter square to kilometre square, divide the area value by 10^-6 ( 0.000001)

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Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to
fenix001 [56]

Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

F = 350 N

Mass = 100kg

350N = 100×a

a = 350/100

a = 3.5m/s^2

The acceleration of the object will be 3.5m/s^2

6 0
3 years ago
(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r
Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface U_1=\frac{GM_em}{R_e}

Gravitational Potential Energy at height h is U_2=\frac{GM_em}{R_e+h}

Energy required to lift the satellite E_1=U_1-U_2

E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

h=\frac{R_e}{2}

h=3.19\times 10^6\ m

(b)For greater height E_1  is greater than E_2

thus energy to lift the satellite is more than orbiting around earth

4 0
3 years ago
Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Give
Keith_Richards [23]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

Answer:

The electric field vector of the satellite broadcast as measured at the surface of the earth is  E_o = 6.995 *10^{-6} \ V/m

Explanation:

From the question we are told that

     The height of the satellite is  r  = 35000 \ km  =  3.5*10^{7} \ m

      The power output of the satellite is P  = 1 \ KW  =  1000 \ W

       

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is  mathematically represented as  

     I  =  \frac{P}{4 \pi r^2}

substituting values

      I  =  \frac{1000}{4 * 3.142 (3.5*10^{7})^2}

      I  = 6.495*10^{-14} \  W/m^2

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be   mathematically represented as  

          I  =  c * \epsilon_o * E_o^2

Where E_o is the amplitude of the electric field vector of the satellite broadcast so

         E_o =  \sqrt{\frac{2 * I}{c * \epsilon _o} }

substituting values

          E_o =  \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }

           E_o = 6.995 *10^{-6} \ V/m

 

   

4 0
3 years ago
The specific heat capacity of sea water is 4100 J/Kg°C and the boiling point of 100.6 °C. (i) Calculate the energy required to r
maw [93]

Answer:

334.314 (kJ)

Explanation:

1) the formula for the required energy is: Q=c*m(Bp-t), where c - 4100 J/kg*C; m - 0.9 kg; Bp - 100.6 C; t - 10 C.

2) according to the formula above:

Q=4100*0.9*(100.6-10)=41*9*906=334314 (J).

6 0
3 years ago
If a marble is released from a height of 10 meters how long would it take to hit the ground?
zlopas [31]
S = ut + 0.5at^2 

<span>10 = 0 + 0.5(9.81)t^2 {and if g = 10 then t^2 = 2 so t ~1.414}  </span>

<span>t^2 ~ 2.04 </span>

<span>t ~ 1.43 seconds</span>
3 0
3 years ago
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