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igomit [66]
3 years ago
7

A helicopter lifts a 66 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/

11. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?
Physics
1 answer:
artcher [175]3 years ago
4 0

Answer

given,

mass of the astronaut = 66 kg

height of the lift = 15 m

acceleration = g/11 = 0.89 m/s²

a) the work done by helicopter

W = F× h

    = m (a+g)h

    = 66 × (9.8+0.89)15

  W =10583.1 J

b) work done by the gravity

W= mgh

W = - 66 × 9.8 × 15

W = -9702 J

c) Δ KE = net work done

  Δ KE = 10583.1 - 9702

  Δ KE = 881.1 J

d) speed

\dfrac{1}{2}mv^2 = 881.1

v = \sqrt{\dfrac{2 \times 881.1}{66}}

v =5.17 m/s

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Artyom0805 [142]

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1 year ago
A pendulum takes 10 seconds to swing through 2 complete cycles what is it’s period and frequency
Orlov [11]

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7 0
3 years ago
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Science If you want to find the energy quantum of light, you multiply the frequency of the radiation (v) by "h". What is "h"?
weqwewe [10]

Answer:

"h" signifies Planck's constant

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In the equation energy E = h X v

The "h" there signifies Planck's constant

Planck's constant is a value, that shows the rate at which the energy of a photon increases/decreases, as the frequency of its electromagnetic wave changes.

It was named after Max Planck who discovered this unique relationship between the energy of a light wave and its frequency.

Planck's constant, "h" is usually expressed in Joules second

Planck's constant = 6.62607015 \times 10^{-34}  J.s

7 0
2 years ago
A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away
Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d)  Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it.  Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball.   Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.  

   Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

                               y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

                              0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

                               V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

                               x = x(0) + V_i,x*t_c

                               D = 0 + V_i,x*t_c

                               V_o,x + v_r = D/t_c

                                V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

                           V_o = sqrt ( V_o,x^2 + V_o,y^2)

                          V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

       

- The angle with which it should be thrown can be evaluated by trigonometric relation:

                            tan(Q) = ( V_o,y / V_o,x )

                            tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

                                   Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

                           

                               

6 0
2 years ago
An astronaut holds a rock 100m above the surface of Planet X. The rock is then thrown upward with a speed of 15m/s, as shown in
Harlamova29_29 [7]

The acceleration due to gravity of the planet X is 1 m/s².

The given parameters;

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  • initial velocity of the rock, u = 15 m/s
  • time of motion of the rock, t = 10 s

The acceleration due to gravity is calculated as follows;

h = ut - \frac{1}{2} gt^2\\\\100 = 15(10) - (0.5\times 10^2)g\\\\100 = 150 - 50g\\\\50g = 150-100\\\\50g = 50\\\\g = 1 \ m/s^2

Thus, the acceleration due to gravity of the planet X is 1 m/s²

Learn more here: brainly.com/question/24564606

7 0
2 years ago
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