Ask question

Ask question

All categories

3 years ago

5

An elevator (mass 4100 kg) is to be designed so that the maximum acceleration is 0.0400g. what is the maximum force the motor sh

ould exert on the supporting cable? incorrect: your answer is incorrect. n what is the minimum force the motor should exert on the supporting cable?

1 answer:

melomori [17]3 years ago

7 0

Answer:

The maximum force on the supporting cable is 80688 N.

The minimum force on the supporting cable is -164 N.

Explanation:

For maximum force movement of elevator is in upward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a is acceleration of elevator

g is acceleration due to gravity

T is the maximum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 0.04g + 9.8)

T = 80688 N

This is the maximum force on the supporting cable.

For minimum force movement of elevator is in downward direction. Thus, equation of motion is given by,

ma = T - mg

where m is the mass of elevator

a= -0.04g is acceleration of elevator because elevator is moving downward

g is acceleration due to gravity

T is the minimum tension in the supporting cable

T = ma + mg

T = m (a + g)

T = 4100 ( 9.8 - 0.04g)

T = -164 N

This is the minimum force on the supporting cable.

You might be interested in

The water line from the street to my house is 1 inch diameter and made of PVC (i.e. smooth). The line is roughly 450 ft long. Th

jekas [21]

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

= $130\times 6.894$

= $896.22 \ Kpa$

or,

= $896.22\times 10^3 \ Pa$

Z₂ = 10ft

= 3.05 m

$\delta$ = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒ $\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2$

On substituting the values, we get

⇒ $\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2$

⇒ $\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05$

⇒ $P_2=866330 \ P_a$

i.e.,

⇒ $=866330\times 0.000145$

⇒ $=126 \ Psi$

8 0

2 years ago

A reference point allows you to determine the _____________ of an object.

marin [14]

Answer:

a reference point allows you to determine the motion of an object :)

example: when your in a car and u look out the window and your moving fast than the car next to you. It seems that way because the car next to you is actually moving much slower.

8 0

2 years ago

A 65 kg student sits 3 m away from a 70 kg student. What is the magnitude of the gravitational force between the two students?

LiRa [457]

Answer:

$F=3.37\times 10^{-8}\ N$

Explanation:

Given that,

Mass of student 1, m₁ = 65 kg

Mass of student 2, m₂ = 70 kg

The distance between the students, d = 3 m

We need to find the magnitude of the gravitational force between the two students. The formula for the magnitude of the gravitational force between two masses is given by :

$F=G\dfrac{m_1m_2}{d^2}\\\\F=6.67\times 10^{-11}\times \dfrac{65\times 70}{(3)^2}\\F=3.37\times 10^{-8}\ N$

So, the gravitational force between the two students is $3.37\times 10^{-8}\ N$ .

3 0

2 years ago

Both natural processes and human activity are causing some of Earth's permanent

wlad13 [49]

Answer:

The water will get more warm fresh water

Explanation:

8 0

1 year ago

A typical car has 16 L of liquid coolant circulating at a temperature of 95 ∘C through the engine's cooling system. Assume that,

amm1812

Answer:

$\Delta V=0.0592\ L$

Explanation:

Given:

Initial temperature of the coolant, $T_f=95^{\circ}C$

final temperature of the coolant, $T_f=105^{\circ}C$

total volume of the coolant, $V_c=16\ L$

coefficient of volume expansion for coolant, $\beta_c=410\times 10^{-6}\ ^{\circ}C^{-1}$

volume of Al radiator, $V_a=2\ L$

volume of steel radiator, $V_s=14\ L$

<u><em>We have:</em></u>

coefficient of volume expansion for Aluminium, $\beta_a=75\times 10^{-6}\ ^{\circ}C^{-1}$

coefficient of volume expansion for steel, $\beta_a=35\times 10^{-6}\ ^{\circ}C^{-1}$

<u>Now, change in volume of the coolant after temperature rises:</u>

$\Delta V_c=V_c.\beta_c.\Delta T$

$\Delta V_c=16\times 410\times 10^{-6}\times (105-95)$

$\Delta V_c=0.0656\ L$

<u>Now, volumetric expansion in Aluminium radiant:</u>

$\Delta V_a=V_a.\beta_a.\Delta T$

$\Delta V_a=2\times 75\times 10^{-6}\times (105-95)$

$\Delta V_a=0.0015\ L$

<u>Now, volumetric expansion in steel radiant:</u>

$\Delta V_s=V_s.\beta_s.\Delta T$

$\Delta V_s=14\times 35\times 10^{-6}\times (105-95)$

$\Delta V_s=0.0049\ L$

∴Total extra accommodation volume created after the expansion:

$V_X=\Delta V_s+\Delta V_a$

$V_X=0.0049+0.0015$

$V_X=0.0064\ L$

Hence, the volume that will overflow into the small reservoir will be the volume of coolant that will be extra after the expanded accommodation in the radiator.

$\Delta V=\Delta V_c-\Delta V_X$

$\Delta V=0.0656-0.0064$

$\Delta V=0.0592\ L$

3 0

3 years ago