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Travka [436]
3 years ago
11

A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/

10. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her
Physics
1 answer:
postnew [5]3 years ago
8 0

Answer:

(a) 1058.4 J

(b) -10584 J

Explanation:

Parameters given:

Mass of astronaut, m = 72 kg

Distance moved by astronaut, d = 15 m

(a) WORK DONE BY FORCE FROM THE HELICOPTER

Work done is given as the product of Force applied to a body and the distance moved by the body:

W = F * d

The force from the helicopter is given as:

F = m * a

where a = acceleration of the astronaut due to the helicopter

Therefore, the work done is given as:

W = m * a * d

W = 72 * g/10 * 15

W = \frac{72 * 9.8 * 15}{10}

W = 1058.4 J

(b) WORK DONE BY FORCE OF GRAVITY

W = F * d

The force of gravity is given as:

F = -m * g

where g = acceleration due to gravity

The negative sign is due to the fact that the astronaut moves in an opposite direction (upwards) to the force of gravity (Gravity acts downwards)

Therefore, the work done is given as:

W = -m * g * d

W = -72 * 9.8 * 15

W = -10584 J

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Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to
sergij07 [2.7K]

Answer:

KE=1.2036\times 10^{-12}\ J

Explanation:

Given:

  • charge on the alpha particle, q=2e=3.2\times 10^{-19}\ C
  • mass of the alpha particle, m=6.64\times 10^{-27}\ kg
  • strength of a uniform magnetic field, B=0.5\ T
  • radius of the final orbit, r=0.5\ m

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

q.v.B=m.\frac{v^2}{r}

m.v=q.B.r

where:

v = velocity of the alpha particle

v=\frac{q.B.r}{m}

v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}

v=1.2048\times 10^{7}\ m.s^{-1}

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m

m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }

m'=6.6533\times10^{-27}\ kg

now kinetic energy:

KE=m'.c-m.c

KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2

KE=1.2036\times 10^{-12}\ J

6 0
3 years ago
Think about the way optical fibers are designed. why are the ends of the fibers glowing very brightly while the sides of the fib
Effectus [21]

Answer:

D. because the light is reflected back into the fiber along its sides

Explanation:

The fiber is constructed in a way that the light is bent/reflected/refracted toward the center core of glass. So, from the center core, there is a layer above it that has a different propagation than the core, and above that the same thing. To give you a real world visual example, if you look down in a pool of water, then stick a straight stick into it, you see that the straight stick appears to bend. That is what is happening to the light as it travels through a different medium (air to water). This same effect is incorporated in the fiber optic cable construction.

6 0
3 years ago
Crest : trough :: compression : _____ A. frequency B. amplitude C. rarefaction D. wavelength
9966 [12]
Rarefraction.

Crest- tallest spot on transverse wave.

Trough- shortest point on transverse wave.

Compression - spot on a compressional wave where the waves are closer together.

Rarefraction - spot on a compressional wave where the waves are farther apart.
3 0
3 years ago
In which situation is work not being done?
almond37 [142]

AS

work done =W = F.d = F d cosФ     (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body.  F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0

Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0

example:  A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.

The third term upon which work done  dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0   ( as cos 90°=0)

5 0
3 years ago
How many nanoseconds does it take light to travel 3.50 ft in vacuum?
Fiesta28 [93]
Answer:3.56 nanosecond

In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:

time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft=  3.56 10^{-9} second or 3.56 nanosecond
6 0
3 years ago
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