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astraxan [27]
3 years ago
13

The international space station (ISS) orbits the earth with a velocity 7.6 km/s. How much energy is required to lift a payload o

f 2500 kg from the surface of the earth to the ISS
Physics
1 answer:
Dmitry [639]3 years ago
4 0

Answer:

7.22×10¹⁰ J

Explanation:

From the question given above, the following data were obtained:

Velocity (v) = 7.6 Km/s

Mass (m) = 2500 kg

Energy (E) =?

Next, we shall convert 7.6 Km/s to m/s. This can be obtained as follow:

1 km/s = 10³ m/s

Therefore,

7.6 Km/s = 7.6 Km/s × 10³ m/s / 1 km/s

7.6 Km/s = 7.6×10³ m/s

Finally, we shall determine the energy required as follow:

Velocity (v) = 7.6×10³ m/s

Mass (m) = 2500 kg

Energy (E) =?

E = ½mv²

E = ½ × 2500 × (7.6×10³)²

E = 1250 × 57760000

E = 7.22×10¹⁰ J

Thus, 7.22×10¹⁰ J of energy is required.

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The kinetic energy of the proton is 3.4 kev
1 kev = 1.602 × 10^-16 joules
therefore 3.4 kev is equivalent to;
3.4 ×  (1.602 ×10^-16)= 5.4468 × 10^-16 J
Kinetic energy is calculated by the formula 1/2mv² where m is the mass and v is the velocity.
Therefore V = √((2 × ( 5.4468×10^-16))/ (1.67 ×10^-27))
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The tub of a washer goes into its spin-dry cy-cle, starting from rest and reaching an angularspeed of 3.1 rev/s in 10.7 s.At thi
neonofarm [45]

Answer:

35 revolutions

Explanation:

t = Time taken

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

\theta = Number of rotation

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{3-0}{10.7}\\\Rightarrow \alpha=0.28037\ rev/s^2

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{3.1^2-0^2}{2\times 0.28037}\\\Rightarrow \theta=17.13806\ rev

Number of revolutions in the 10.7 seconds is 17.13806

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-3.1}{11.2}\\\Rightarrow a=-0.27678\ rev/s^2

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-3.1^2}{2\times -0.27678}\\\Rightarrow \theta=17.36035\ rev

Number of revolutions in the 11.2 seconds is 17.36035

Total total number of revolutions in the 21.9 second interval is 17.13806+17.36035 = 34.49841 = 35 revolutions

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4 years ago
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