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QveST [7]
3 years ago
9

People hoping to travel to other worlds are faced with huge challenges. One of the biggest is the time required for a journey. T

he nearest star is 4.1×1016m 4.1 × 10 16 m away. Suppose you had a spacecraft that could accelerate at 1.5 g g for two thirds of a year, then continue at a constant speed. (This is far beyond what can be achieved with any known technology.)How long would it take you to reach the nearest star to earth?
Physics
1 answer:
Elena L [17]3 years ago
5 0

Answer:

It would take 8.22037 hrs away. Wouldn't it?

Explanation:

Because

    4.11016

    4.11016

            15

= 8.22037

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A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (
natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

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If ball 4 has a mass of 2 kg and it is 5m high, what will be its gravitational potential energy? (g=10 N/kg) *
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Explanation:

Gravitational potential energy

= mgh

= (2kg)(10N/kg)(5m)

= 100J.

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Is a stick living,non living or Dead ?
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A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
spayn [35]

The magnitude of the electric field can be calculated using the equation

E = \frac{F}{q}, where F is the Force acting on the proton and q is the charge of the proton.

We know that the charge of the proton is q = 1.6 x 10^{-19}  C

We have to calculate the Force first.

We know that F = ma from Newton's 2nd law, where m is the mass of the proton and a is the acceleration.

We know that the mass of the proton m = (1.67) X 10^{-27}  kg

So it turns out that we have to calculate acceleration before anything else.

In order to calculate the acceleration, we make use of the following data from the question:

Initial Velocity of the proton V_{i}  = 0

Distance traveled  D = 1.70 cm = 0.017 m

Time taken for the travel between the plates t = (1.48) X 10^{-6}  s

Acceleration a = ?

Using the equation, D = V_{i}t + \frac{1}{2} at^{2}, we get

Knowing that initial velocity is 0, the equation reduces to D = \frac{1}{2}at^{2}

Rearranging the equation so as to make a the subject of the formula, we have

a = \frac{2D}{t^{2} }

Plugging in the numbers and simplifying gives us a = 1.5 x 10^{10}   m/s^{2}

We can now calculate the Force using F = ma

Plugging in the known values, we get F = 2.5 x 10^{-17}  N

Using this, we can calculate E through the equation E = \frac{F}{q}

Plugging the numbers and simplifying gets us E = 156.25 N/C

Thus, the magnitude of the electric field between the plates of the capacitor is 156.25 N/C


B) To calculate the Final Velocity of the proton, we can make use of the equation

V_{f}  = V_{i}  + at

Plugging the numbers in and simplifying gets us V_{f}  = (2.22)  *  10^{4}  m/s

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