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ASHA 777 [7]
3 years ago
5

Stan does 178 J of work lifting himself 0.5 m. What is Stan’s mass? The acceleration of gravity is 9.8 m/s 2 . Answer in units o

f kg
Physics
2 answers:
svet-max [94.6K]3 years ago
8 0

Answer:

<h2>The mass is 36.33 kilograms.</h2>

Explanation:

The work is defined as the force needed to move an object a certain distance, if there's no change of position, there's no work.

In this case, the work is defined:

W=Fd

Where F is the weight defined with the following equation

F=W=mg

Now, we'll replace all given values and solve for m:

W=mgy\\178=m(9.8)(0.5)\\m=\frac{178}{4.9}=36.33 kg

Therefore, the mass is 36.33 kilograms.

Rom4ik [11]3 years ago
4 0
Work= (force)(distance)
178= m(9.81)x0.5
178=m(4.905)
178/4.905=m

His mass is 36.3 kg
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Answer:

t=0.038s

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Rearrange this equation to find time t

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t=m\frac{v_{f}-v_{i}}{F}

Substitute the given values

t=3.8kg\frac{9.3*10^3m/s-0}{9.3*10^5N} \\t=0.038s  

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3 years ago
A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
Setler [38]

Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

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speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

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(a) The wavelength of the sound between source and the listener is given by:

\lambda = \frac{v_{a}}{\vartheta }         (2)

(b) The observed frequency is given by:

\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

\vartheta = \frac{334}{334 - 80}\vartheta_{o}

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\lambda = 0.7604\lambda_{o}

4 0
3 years ago
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monitta

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A football kicker kicks a ball with a mass of .42 kg. The average acceleration of the football was 14.8 m/s squared. How much fo
mrs_skeptik [129]

To find the force we use the formula,

F = ma , where m is mass and a acceleration

Using the formula,

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2 years ago
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Answer:

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Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

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for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

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taking the square root this gives us :

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5 0
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