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3 years ago

Activity will have the LEAST effect on your level of physical fitness?

Physics

1 answer:

Scrat [10]3 years ago

6 0

Answer:

The correct answer is option D, Using an Electric mixer ( i hope this is the correct answer to what you are talking about because you didn't have any options but i gave out almost the same question to my students.)

Explanation:

Physical fitness is assured by activities in which our body is doing some kind of physical movements or exercise. In case of “walking to work “ the body is compelled to move or walk, in case of “carrying boxes up and down stairs” lot of physical work is done in both carrying the boxes and moving up and down the stairs, in the same way in case of “shoveling snow” some physical efforts are required to shovel the snow. But in case of using a electric mixer, the only work being done is the “pressing of ON/OFF button”. Thus , "using an electric mixer" will have the LEAST effect on your level of physical fitness

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9. 2076 Set B Q.No. 9b Two galvanometers, which are otherwise identical; are fitted with different coils. One has coil of 50 tur

Hunter-Best [27]

The ratio of the deflection when each is connected in turns to a cell of e.m.f 25 V and internal resistance 50 ohms is 13: 12

<h3></h3><h3>What is internal resistance?</h3>

Internal resistance can be described as the resistance within a battery, or other voltage source, that causes a drop in the source voltage when there is a current.

The parameters given are :

Coil 1 = 50 turns

Coil 2 = 500 turns

Resistance 1 = 1022

Resistance 2 = 6002

Internal resistance = 50 ohms

Emf = 25v

I = 25/ 50+ 10

I = 25/60

I = 5/ 12 A

$I_{2}$ = 25/50+ 600

= 25/ 650

= 5/ 130 A

The ratio of the deflection when each is connected in turns to a cell of e.m. 25 V and internal resistance 50 ohms =

Q1/ Q2

Q1 = N1 x B x $I_{1}$ / c

Q2 = N2 x B x $I_{2}$ / c

therefore Q/ Q2 = (50 x 5/12 )/ (500 x 5/130)

Q1 / Q2 = 12/ 13 which 13: 12

Therefore, The ratio of the deflection when each is connected in turns to a cell of e.m. 25 V and internal resistance 50 ohms is 13: 12.

Learn more about internal resistance at; brainly.com/question/20595977

#SPJ1

6 0

1 year ago

How does an airplane engine work?

vivado [14]

Answer:

Explanation:

Un motor aeronáutico o motor de aviación es aquel que se utiliza para la propulsión de aeronaves mediante la generación de una fuerza de arrastre.

Existen distintos tipos de motores de aviación, aunque se dividen en dos clases básicas: motores recíprocos —o de pistón— y de turbina de gas. Recientemente y gracias al desarrollo de la NASA y otras entidades, se ha comenzado también la producción de motores eléctricos para aeronaves que funcionen con energía solar fotovoltaica.

eso depende de el motor que cada avion tenga

7 0

3 years ago

10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.

Xelga [282]

Answer:

$\lambda=1282\ nm$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

So,

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

Given, $n_i=5\ and\ n_f=3$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m$

$\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m$

$\lambda= 1.2828\times10^{-6}$

1 m = 10⁻⁹ nm

$\lambda=1282\ nm$

6 0

3 years ago

A 37.5 kg box initially at rest is pushed 4.05 m along a rough, horizontal floor with a constant applied horizontal force of 150

vodomira [7]

Answer:

a) 607.5 J

b) 160.531875 J

c) 0 J

d) 0 J

e) 2.925 m\s

Explanation:

The given data :-

Mass of the box ( m ) = 37.5 kg.

Displacement made by box ( x ) = 4.05 m.

Horizontal force ( F ) = 150 N.
The co-efficient of friction between box and floor ( μ ) = 0.3
Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b) The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the ( K.E )₁ = 0

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d) The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.

5 0

4 years ago

Please answer the following question!

Semmy [17]

Answer:

Explanation:

the momentum will always be 0 when it is at rest because the object isnt moving!

Hope this helped!

6 0

2 years ago