Ask question

Ask question

All categories

sertanlavr [38]

3 years ago

6

(a) what is the acceleration of two falling sky divers (mass 132 kg including parachute) when the upward force of air resistance

is equal to one-fourth of their weight? (b) after popping open the parachute, the divers descend leisurely to the ground at constant speed. what now is the force of air resistance on the sky divers and their parachute?

1 answer:

Komok [63]3 years ago

4 0

As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity F=W= mg [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

=132 kg ×9.8 m/s^2

=1293.6 kg m/s^2

=1293.6 N [ here N[newton] is the unit of force.]

As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

$=\frac{1}{4} *1293.6N$

$=323.4 N$

Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

$F_{net} =1293.6N -323.4N$

$=970.2 N$

From Newton's second law of motion we know that net force is the product of mass and acceleration i.e

$F_{net} =ma$ [Here a is the acceleration]

$a =\frac{F_{net} }{m}$

$= \frac{970.2}{132} m/s^2$

$=7.35 m/s^2$

In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

we know that F= ma

=m×0

=0 N

Hence they will not get any force when they will descend with a uniform speed.

You might be interested in

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cbold%7B%20Choose%20%5C%3A%20the%20%5C%3A%20correct%20%5C%3A%20Answer%20%5C%3A%20%3A%29%20%5

Ulleksa [173]

Answer:

B and B

Explanation:

Question 1:

The distance is a unit of length, such as miles, meters, kilometers

The correct answer for this question would be 8m

Question 2:

Speed has a unit of length per unit of time.

The correct answer for this would be meters per second

-Chetan K

3 0

2 years ago

Two charges repel each other with a force. If the magnitude of both charges double, but the distance between them remains consta

-Dominant- [34]

Answer:

C.

Explanation:

4 0

3 years ago

A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t

Leona [35]

Answer:

To obtain the power, we first need to find the work made by the force.

1) To calculate the work, we need the next equation:

$\int\limits {F} \, dx$

So the force is given by the problem so our mission is to find 'dx' in terms of 't'

2) we know that:

$\frac{dV}{dt} = a = 2.6$

So we have:

$v = 2.6t$

Then:

$\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt$

3) Finally, we replace everything:

$\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt$

After some calculation, we have as a result that the work is:

161.9638 J.

4) To calculate the power we need the next equation:

$P = \frac{W}{t}$

So

P = 161.9638/4.7 = 34.46 W

8 0

3 years ago

Just this last one!!

Pepsi [2]

Answer:

$47 \ \frac{m}{s}$

Explanation:

s = displacement (m)

u = initial velocity $(\frac{m}{s})$

v = final velocity $(\frac{m}{s})$

a = acceleration $(\frac{m}{s^{2} })$

t = time (s)

s = 235

a = -4.7

v = 0

v² = u² + 2as

(0)² = u² + 2(-4.7)(235)

u² - 2209 = 0

u² = 2209

u = 47

4 0

2 years ago

Read 2 more answers