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3 years ago

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Wyatt is moving a box with a mass of 37 kg a distance of 37 meters. Wyatt did 360 J of work in 2 minutes when moving the box. Wh

at was his power output? A) 9.72 W B) 3 W C) 180 W D) 323 W

2 answers:

pentagon [3]3 years ago

7 0

His power output was 3 Watt (360 Joule/120 seconds). The power output can be calculated by dividing the quantity of work by the amount of second needed for the activity and also by multiplying the force amount with the velocity of the activity. The power output usually used for measuring the ability of machine for doing its job.

enot [183]3 years ago

4 0

Answer:

Power output, P = 3 watts

Explanation:

It is given that,

Work done by Wyatt, W = 360 J

Time taken, t = 2 minutes = 120 s

We need to find the power output. It is given by:

$P=\dfrac{W}{t}$

Where

W = work done

t = time taken

So, $P=\dfrac{360\ J}{120\ s}$

P = 3 Watts

So, the power output is 3 to watts. Hence, this is the required solution.

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A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi

Darina [25.2K]

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

5 0

3 years ago

Calcium oxide write this in a formula

kykrilka [37]

The formula for calcium oxide is CaO.

4 0

3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

A 1.00 -kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Fig. P8.62a). The object has

LekaFEV [45]

The distance D where the object comes to rest is 1.08.m.

<h3>What is the distance?</h3>

The separation of one thing from another in space; the distance or separation in space between two objects, points, lines, etc.; remoteness. The distance of seven miles cannot be accomplished in one hour of walking.
Learn how to use the Pythagorean theorem to get the separation between two points using the distance formula. The Pythagorean theorem can be rewritten as d==(((x 2-x 1)2+(y 2-y 1)2)
The distance between any two places is the length of the line segment separating them. By measuring the length of the line segment that connects the two points in coordinate geometry, the distance between them may be calculated.

(c) the distance D where the object comes to rest.

$W_{total} =$ ΔKE ⇒ -0.25*1*9.8*D = 0-1/2*1* $2.3^{2}$

⇒ $D=\frac{0.5*2.3^{2} }{2.45}$

⇒1.08.m

To learn more about distance, refer to:

brainly.com/question/4998732

#SPJ4

5 0

1 year ago

For thermal equilibrium at temperature Tan appropriate measure of energy is kT where k is Boltzmann's constant. Convert the foll

Schach [20]

Answer:

1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K

Explanation:

The relationship between temperature and thermal energy is

E = K T

The relationship of the speed of light

c =λ f = f / ν 1/λ= ν

The Planck equation is

E = h f

Let's start the transformations

c = f λ = f / ν

f = c ν

E = h f

E = h c ν

E = KT

h c ν = K T

T = h c ν / K =( h c / K) ν

Let's replace the constants

h = 6.63 10⁻³⁴ J s

c = 3 10⁸ m / s

K = 1.38 10⁻²³ J / K

v = 1 cm-1 (100 cm / 1 m) = 10² m-1

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²

A = h c / K = 1,441 10⁻²

T = 1.44K

ν = 103 cm⁻¹ = 103 10² m

T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²

T = 148K

1 Rydberg = 1.097 10 7 m

As we saw at the beginning the λ=1 / v

T = (h c / K) 1 /λ

T = 1,441 10⁻² 1 / 1,097 10⁷

T = 1.3 10⁻⁹ K

E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J

E = KT

T = E/K

T = 1.6 10⁻¹⁹ /1.38 10⁻²³

T = 1.16 10⁴ K

3 0

3 years ago