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astraxan [27]
2 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]2 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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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
3 years ago
a force is applied to a block causing it to accelerate along a horizontal, frictionless surfaces. the energy gained by the block
GaryK [48]
Energy given to the block
3 0
3 years ago
What connects the upper motor neurons to lower motor neurons?
skad [1K]

Answer:

The upper motor neurons synapse in the spinal cord connect with anterior horn cells of lower motor neurons, usually via interneurons. The anterior horn cells are the cell bodies of the lower motor neurons and are located in the grey matter of the spinal cord.

Explanation:

Interneurons are the central nodes of neural circuits, enabling communication between the upper motor neurons, sensory or motor neurons located in the brain and spinal cord and they send signals to lower motor neurons or central nervous system (CNS) in the brain stem and spinal cord . When they get a signal from the upper motor neurons, they send another signal to your muscles to make them contract. They play vital roles in reflexes, neuronal oscillations, and neurogenesis in the adult mammalian brain.

Renshaw cells are among the very first identified interneurons. They are excited by the axon collaterals of the motor neurons. In addition, Renshaw cells make inhibitory connections to several groups of motor neurons.

7 0
3 years ago
A hammer strikes one end of a thick iron rail of length 8.80 m. A microphone located at the opposite end of the rail detects two
stepladder [879]

Answer:

ΔT = 0.02412 s

Explanation:

We will simply calculate the time for both the waves to travel through rail distance.

FOR THE TRAVELING THROUGH RAIL:

T_{rail} = \frac{Distance}{Speed\ of\ Sound\ in\ Rail}\\\\T_{rail} = \frac{8.8\ m}{5950\ m/s}\\\\T_{rail} = 0.00148\ s

FOR THE WAVE TRAVELING THROUGH AIR:

T_{air} = \frac{Distance}{Speed\ of\ Sound\ in\ Air}\\\\T_{air} = \frac{8.8\ m}{343\ m/s}\\\\T_{air} = 0.0256\ s

The separation in time between two pulses can now be given as follows:

\Delta T = T_{air}-T_{rail} \\\Delta T = 0.0256\ s - 0.00148\ s\\

<u>ΔT = 0.02412 s</u>

3 0
3 years ago
What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t
Gekata [30.6K]

Start by facing East. Your first displacement is the vector

<em>d</em>₁ = (225 m) <em>i</em>

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

<em>d</em>₂ = (350 m) <em>j</em>

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )

<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>

The net displacement is

<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃

<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>

and its magnitude is

|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0
3 years ago
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