The higher an object is raised compared to the ground

How will you join three resistances, each of 2 ohm so that the effective resistance is 3 ohm?

svet-max [94.6K]

First I will parallel two of the resistors, creating a net 1 ohm. Then I will series that with the remaining 2-ohm resistor, resulting in 3 ohms.

5 0

3 years ago

3) A dock worker pushes a 72 kg crate up a 2.0 m high,

Vlad [161]

Work done on the crate is 1411.2 J

Explanation:

Work done is defined as the product of force and the distance moved by the object. The unit of work done is in joules and denoted by the symbol J.

Work done = F * d

where F represents the force and d represents the distance moved by the object.

mass = 72 kg , distance moved by the object is given by 2.0 m

Force F = mass * gravity = 72 * 9.8

= 705.6 N =706 N.

Work done = 706 * 2.0 = 1412 J.

7 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

Island arcs are a product of _________ whereas mountain ranges are a product of _________. Volcanic

Nadusha1986 [10]

Answer:

c) Subduction of an oceanic plate under another oceanic plate, collision of two continental plates, subduction of an oceanic plate under a continental plate

Explanation:

Islands arcs are a product of subduction of an oceanic plate under another oceanic plate

Mountain ranges are a product of collision of two continental

plates,

Volcanic arcs are a product of subduction of an oceanic plate under a continental plate

4 0

3 years ago

In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape

lina2011 [118]

Answer:

The deceleration of the pilot was 1.9 x 10⁴ m/s²

Explanation:

First, let´s calculate the velocity of the pilot 3 m above the ground. To do that, we need to know how much time the pilot was falling. The equation for the position and velocity of the pilot are as follows:

y = y0 + v0 * t + 1/2 * g * t²

v = v0 + g * t

where

y = position of the pilot at time t

y0 = initial position

v0 = initial velocity

t = time

v = velocity at time t

g = acceleration due to gravity

If we consider the ground as the center of our reference system and that the pilot fell with an initial velocity of 0 (the pilot would unlikely impulse himself to the ground), then the equations would be as follows:

y = y0 + 1/2 g * t²

v = g * t

The time at which the pilot was 3.0 m above the ground will be:

3.0 m = 6000 m - 1/2 * 9.8 m/s² * t²

3.0 m - 6000 m / -4.9 m/s² = t²

t = 35.0 s

The velocity at that time will be:

v = -9.8 m/s² * 35.0 s = -343 m/s

After 35.0 s the pilot has a positive acceleration besides the acceleration due to gravity. Then, the equation for velocity and position will be:

v = v0 + a*t now, v0 = -343 m/s and a ≠ g and a>0

y = y0 + v0 * t + 1/2 * a * t² now, y0 = 3 m

Again, let´s find the time at which the pilot hits the ground:

v = v0 + a*t

v-v0/ t = a

Replacing in the equation for position:

y = y0 + v0 * t + 1/2 * ((v-v0)/t) * t²

y = y0 + v0 * t + 1/2 * v* t - 1/2 * v0 * t)

y = y0 + 1/2 v0 * t + 1/2 v * t

replacing with numbers:

0m = 3m + 1/2 * (-343 m/s)t + 1/2 *(-54 m/s)t

-3.0 m = - 198.5 m/s * t

t = -3m / -198.5 m/s

t =0.015 s

the upward acceleration was then:

v-v0/ t = a

-54 m/s -(-343 m/s) / 0.015 s = a

a = 1.9 x 10⁴ m/s²

6 0

3 years ago