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LenaWriter [7]
3 years ago
7

Iron cutting shers has a short cutting edge ,whereas a pair of cloth cutting scissors has a long a long cutting edge​

Physics
1 answer:
Travka [436]3 years ago
6 0

cloth cutting scissors have long cutting edge because to cut the clothes easily

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The carpenter question plz ASAP
k0ka [10]

Answer:

d. The hammer falls with a constant acceleration

Explanation:

Since gravity is the only thing that is acting on the hammer as it falls and gravity is a form of acceleration then acceleration of 9.81m/s² which is gravity is the correct answer.

6 0
3 years ago
How much work is done when you push a crate horizontally with 130 N across a 5.0-m
erastova [34]

Complete solution is given below in attachment

8 0
3 years ago
The majority of which type of dancing was created in African American communities?
matrenka [14]

Answer:

swing dances

Explanation:

8 0
2 years ago
Read 2 more answers
A horizontal 745 N merry-go-round of radius
Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

m = \frac{ weight}{g}

m = \frac{745}{9.81 }

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =0.5 \times m \times r^2

Substituting the values

Moment of Inertia of solid cylinder I  

=>0.5 \times 76.02 \times (1.45)^2

=> 0.5 \times 76.02\times 2.1025

=> 79.91 kg.m^2

Step 3: Finding the Torque applied T

Torque applied T = F \times r

Substituting the values

T = 56.3  \times 1.45

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,\alpha  = \frac{Torque}{Inertia}

Substituting the values,

\alpha  = \frac{81.635}{79.91}

\alpha = 1.021 rad/s^2

 Step 4: Finding the Final angular velocity

Final angular velocity ,\omega = \alpha \times  t

Substituting the values,

\omega = 1.021 \times  3.62

\omega = 3.69 rad/s

Now KE (100% rotational) after 3.62s is:

KE = 0.5 \times I \times \omega^2

KE =0.5 \times 79.91 \times 3.69^2

KE = 544J

6 0
4 years ago
A monatomic ideal gas initially fills a container of volume V = 0.25 m3 at an initial pressure of P = 250 kPa and temperature T
Murrr4er [49]

Answer:

number of moles = 27.34 moles

the temperature of gas after it undergoes the isobaric expansion = 605 K

Explanation:

Given that:

V = 0.25 m³

P = 250 kPa

T = 275 K

V₂ = 0.55 m³

P₂ = 760 kPa

a)

Using ideal gas equation ; PV = nRT

n = \frac{PV}{RT}\\\\n =  \frac{250*10^3*0.25}{8.314*275}\\\\n = \frac{62500}{2286.35}\\\\n = 27.34 \ moles

b) To calculate the temperature of gas after it undergoes the isobaric expansion; we have:

  1. \frac{V_1}{T_1}= \frac{V_2}{T_2}\\\\\frac{0.25}{275}= \frac{0.55}{T_2}\\\\T_2=\frac{0.55*275}{0.25}\\\\T_2 = 605 K

4 0
4 years ago
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