All categories

3 years ago

How much does a 725 kg elevator weigh? F = [?]N

Physics

2 answers:

olga2289 [7]3 years ago

8 0

725 Kg equals 725x9.8= 7105 N

Assoli18 [71]3 years ago

7 0

Answer:

7105N

Explanation:

725 kg = 725.9.8=7105N

You might be interested in

What time does the clock go back for daylight savings

Novosadov [1.4K]

Answer:

it goes back an hour from the time it was

6 0

2 years ago

What is a boyfriend to you? And what age is appropriate to have one?

Sauron [17]

A boyfriend is someone who you find more than a friend. Someone who you have a deeper connection with. An age to have one really depends on the person and is mostly decided by parents. If the person is mature they may have one around 15 to 16+ years old. If the person is still "immature" or is not ready to have one, the age limit may be extended to 17+

7 0

2 years ago

Read 2 more answers

Neon atoms at 245 K pass through a fan that gives each mole of neon gas an additional kinetic energy of 16.0 J. Part A What is t

hjlf

Answer:

246.28 K

Explanation:

The total energy of one mole of gas molecules can be calculated by the formula given below

E = $\frac{3}{2}\times R\times T$

Where R is gas constant and T is absolute temperature.

Put the value of R as 8.314 and temperature as 245 , we get

E = $\frac{3}{2}\times 8.314\times 245$

= 3055.4 J

Add 16 j to it

Total energy of gas molecules = 3055.4 + 16 = 3071.4 J.

If T be the temperature after addition of energy then

$\frac{3}{2}\times 8.314\times T$ = 3071.4

T = $\frac{2\times 3071.4}{3\times 8.314}$

T = 246.28 K

7 0

3 years ago

If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int

vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

$Q$ is the heat absorbed by the system

$W$ is the work done by the system on the surrounding

In this problem, the work done by the system is

$W=-225 kcal=-941.4 kJ$

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

$Q=-5 \cdot 10^2 kJ=-500 kJ$

with a negative sign as well because it is released by the system.

Therefore, by using the initial equation, we find

$\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ$

8 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago