When not in a vacuum, what force causes objects in freefall to fall at different speeds??

a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while th

anzhelika [568]

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

$F_{1}$ = 250

$A_{1}$ = Area of the small piston = π $r^{2}$

$A_{1}$ = 22/7 x $0.4^{2}$

$A_{1}$ = 0.5 $m^{2}$

$F_{2}$ = ?

$A_{2}$ = Area of the large piston = π $r^{2}$

$A_{2}$ = π x 1

$A_{2}$ = 3.14 $m^{2}$

To calculate the force on the large piston, we will use the below formula

$F_{1}$ / $A_{1}$ = $F_{2}$ / $A_{2}$

Substitute all the parameters into the equation

250/0.5 = $F_{2}$ /3.14

$F_{2}$ = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

$F_{1}$ / 0.25 $A_{1}$ = $F_{2}$ / $A_{2}$ h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

Learn more here: brainly.com/question/13596980

5 0

2 years ago

I REALLY NEED HELP What is a closed physical system A.A system where matter and energy can enter, but not leave the system B.non

NNADVOKAT [17]

You probably already took this but if anyone wanted to know the answer it's D. "A system where matter and energy cannot enter or leave the system".

(I also took the quiz for it and got it correct)

4 0

3 years ago

An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed incre

yarga [219]

Answer:

a. F = 2.32*10^-18 N

b. The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

$v^2=v_o^2+2ax$ (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

$a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}$

Next, you use the second Newton law to calculate the force:

$F=ma$

m: mass of the electron = 9.11*10^-31kg

$F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N$

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

$F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N$

The quotient between the weight of the electron and the force F is:

$\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}$

The force F is 2.59*10^11 times the weight of the electron

8 0

3 years ago

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So