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bonufazy [111]
2 years ago
5

A 10 kg package is delivered to your house. Use one complete sentence to describe an example of how work is done on the package

as it gets brought inside. Make sure to use proper spelling, grammar, and other language mechanics. In your explanation, make sure to use the terms related to the formula for work (W = Fd).
Physics
1 answer:
Mamont248 [21]2 years ago
8 0

Answer:

Hey :)

Explanation:

Work is a net force applied through a distance in order to displace an object, commonly abbreviated as W.  A net force is the sum of all forces acting on an object. Work is mass times acceleration and distance so to find out the work you simply calculate the acceleration of the box being brought in. Next find the distance it was carried to get in the house. Then find out the mas of the box and finally multiply those sums together to get the amount of work put in to bring the package inside.

hope this helps :) xo

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In Physics lab, Ellen has been given the task of constructing a simple motor. She must find common household items at home and b
Aleks [24]

Answer:

a) battery-->electrical current-->copper wire rotor -->magnet--> mechanical energy

Explanation:

6 0
2 years ago
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A.Whale communication. Blue whales apparently communicate with each other using sound of frequency 17.0 Hz, which can be heard n
Y_Kistochka [10]

A. 90.1 m

The wavelength of a wave is given by:

\lambda=\frac{v}{f}

where

v is the speed of the wave

f is its frequency

For the sound emitted by the whale, v = 1531 m/s and f = 17.0 Hz, so the wavelength is

\lambda=\frac{1531 m/s}{17.0 Hz}=90.1 m

B. 102 kHz

We can re-arrange the same equation used previously to solve for the frequency, f:

f=\frac{v}{\lambda}

where for the dolphin:

v = 1531 m/s is the wave speed

\lambda=1.50 cm=0.015 m is the wavelength

Substituting into the equation,

f=\frac{1531 m/s}{0.015 m}=1.02 \cdot 10^5 Hz=102 kHz

C. 13.6 m

Again, the wavelength is given by:

\lambda=\frac{v}{f}

where

v = 340 m/s is the speed of sound in air

f = 25.0 Hz is the frequency of the whistle

Substituting into the equation,

\lambda=\frac{340 m/s}{25.0 Hz}=13.6 m

D. 4.4-8.7 m

Using again the same formula, and using again the speed of sound in air (v=340 m/s), we have:

- Wavelength corresponding to the minimum frequency (f=39.0 Hz):

\lambda=\frac{340 m/s}{39.0 Hz}=8.7 m

- Wavelength corresponding to the maximum frequency (f=78.0 Hz):

\lambda=\frac{340 m/s}{78.0 Hz}=4.4 m

So the range of wavelength is 4.4-8.7 m.

E. 6.2 MHz

In order to have a sharp image, the wavelength of the ultrasound must be 1/4 of the size of the tumor, so

\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4} m

And since the speed of the sound wave is

v = 1550 m/s

The frequency will be

f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz

3 0
2 years ago
A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl
tatuchka [14]

By Newton's second law, assuming <em>F</em> is horizontal,

• the net <u>horizontal</u> force on the <u>larger</u> block is

<em>F</em> - <em>µmg</em> = 3<em>mA</em>

where <em>µmg</em> is the magnitude of friction felt by the larger block due to rubbing with the smaller one, <em>µ</em> is the coefficient of static friction between the two blocks, and <em>A</em> is the block's acceleration;

• the net <u>vertical</u> force on the <u>larger</u> block is

4<em>mg</em> - 3<em>mg</em> - <em>mg</em> = 0

where 4<em>mg</em> is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3<em>mg</em> is the weight of the larger block, and <em>mg</em> is the weight of the smaller block;

• the net <u>horizontal</u> force on the <u>smaller</u> block is

<em>µmg</em> = <em>ma</em>

where <em>µmg</em> is again the friction between the two blocks, but notice that this points in the same direction as <em>F</em>. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net <u>vertical</u> force on the <u>smaller</u> block is

<em>mg</em> - <em>mg</em> = 0

where <em>mg</em> is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for <em>A</em> and <em>a</em> :

<em>A</em> = (<em>F</em> - <em>µmg</em>) / (3<em>m</em>)

<em>a</em> = <em>µg</em>

5 0
1 year ago
Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w
ioda

Answer:

Q_{week} = 458884.6\, BTU

Explanation:

The weekly water consumption of Kimonoski is:

m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{bath.week} = 948.205\,lbm

m_{others, week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (23\,gal)\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{others, week} = 1346.218\,lbm

m_{week} = m_{bath,week} + m_{others, week}

m_{week} = 2294.423\,lbm

The total energy required per week for hot water is:

Q_{week} = m_{week}\cdot c_{p,water}\cdot \Delta T

Q_{week} =(2294.423\,lbm)\cdot (1\,\frac{BTU}{lbm\cdot ^{\textdegree}F} )\cdot (50^{\textdegree}F)

Q_{week} = 458884.6\, BTU

3 0
1 year ago
An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in
vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

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2 years ago
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