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rjkz [21]
3 years ago
12

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? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?
Physics
2 answers:
NemiM [27]3 years ago
8 0

Answer:

a) F_H=776.952\ N

b) F_g=706.32\ N

c) v=5.4249\ m.s^{-1}

d) KE=1059.48\ J

Explanation:

Given:

  • mass of the astronaut, m=72\ kg
  • vertical displacement of the astronaut, h=15\ m
  • acceleration of the astronaut while the lift, a=\frac{g}{10} =0.981\ m.s^{-2}

a)

<u>Now the force of lift by the helicopter:</u>

Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.

F_H-F_g=m.a

where:

  • F_H= force by the helicopter
  • F_g= force of gravity

F_H=72\times 0.981+72\times9.81

F_H=776.952\ N

b)

The gravitational force on the astronaut:

F_g=m.g

F_g=72\times 9.81

F_g=706.32\ N

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, u=0\ m.s^{-1}

using equation of motion:

v^2=u^2+2a.h

v^2=0^2+2\times 0.981\times 15

v=5.4249\ m.s^{-1}

c)

Hence the kinetic energy:

KE=\frac{1}{2} m.v^2

KE=0.5\times 72\times 5.4249^2

KE=1059.48\ J

Scilla [17]3 years ago
7 0

Answer:

Explanation:

mass of helicopter, m = 72 kg

height, h = 15 m

acceleration, a = g/10

(a) Work done by the force

Work, W = force due to helicopter x distance

W = m x ( g + a) x h

W = 72 ( 9.8 + 0.98) x 15

W = 11642.4 J

(b) Work done by the gravitational force

W = - m x g x h

W = - 72 x 9.8 x 15

W = - 10584 J

(c) Kinetic energy = total Work done

K = 11642.4 - 10584

K = 1058.4 J

(d) Let the speed is v.

K = 0.5 x m v²

1058.4 = 0.5 x 72 x v²

v = 5.42 m/s

Now the force of lift by the helicopter:

where:

force by the helicopter

force of gravity

b)

The gravitational force on the astronaut:

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero,

using equation of motion:

c)

Hence the kinetic energy:

Read more on Brainly.com - brainly.com/question/15213128#readmore

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ruslelena [56]

1) No

2) Yes

3) No

4) Equal and opposite

5) 32400 N

6) Repulsive

7) The electric force is 2.3\cdot 10^{39} times bigger than the gravitational force

Explanation:

1)

In nature, the minimum possible charge that an object can have is the charge of the electron, which is called fundamental charge:

e=1.6\cdot 10^{-19}C

Electrons are indivisible particles (they cannot be separated), this means that an object can have at least the charge equal to the charge of one electron (in fact, it cannot have a charge less than e, because it would meant that the object has a "fractional number" of electrons).

In this problem, the object has a charge of

Q=5\cdot 10^{-20}C

If we compare this value to e, we notice that Q, so no object can have a charge of Q.

2)

As we said in part 1), an object should have an integer number of electrons in order to be charged.

This means that the charge of an object must be an integer multiple of the fundamental charge, so we can write it as:

Q=ne

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Q is the charge of the object

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Here we have

Q=2.4\cdot 10^{-18}C

Substituting the value of e, we find n:

n=\frac{Q}{e}=\frac{2.4\cdot 10^{-18}}{1.6\cdot 10^{-19}}=15

n is integer, so this value of the charge is possible.

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We now do the same procedure for the new object in this part, which has a charge of

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Q=ne

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Using the value of the fundamental charge,

e=1.6\cdot 10^{-19}C

We find:

n=\frac{Q}{e}=\frac{2.0\cdot 10^{-19}}{1.6\cdot 10^{-19}}=1.25

n is not integer, so this value of charge is not possible, since an object cannot have a fractional number of electrons.

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To solve this part, we use Newton's third law of motion, which states that:

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In this problem, we have two objects:

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Charge Q exerts an electric force on charge 5Q, and we can call this action force. At the same time, charge 5Q exerts an electric force on charge Q (reaction force), and according to Newton's 3rd law, the two forces are equal and opposite.

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q_2=+7.2\cdot 10^{-6}C is charge 2

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In this problem, the two charges are:

q_1=+4.5\cdot 10^{-6}C is charge 1

q_2=+7.2\cdot 10^{-6}C is charge 2

We see that the two charges have same sign: therefore, the force between them is repulsive.

7)

The electric force between the proton and the electron in the atom can be written as

F_E=k\frac{q_1 q_2}{r^2}

where

q_1 = q_2 = e = 1.6\cdot 10^{-19}C is the magnitude of the charge of the proton and of the electron

r=5.3\cdot 10^{-11} m is the separation between them

So the force can be rewritten as

F_E=\frac{ke^2}{r^2}

The gravitational force between the proton and the electron can be written as

F_G=G\frac{m_p m_e}{r^2}

where

G is the gravitational constant

m_p = 1.67\cdot 10^{-27}kg is the proton mass

m_e=9.11\cdot 10^{-27}kg is the electron mass

Comparing the 2 forces,

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