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ss7ja [257]
3 years ago
10

an object with a mass of 125 kilograms is lifted through a height of 10 meters. how much work is done?

Physics
1 answer:
marissa [1.9K]3 years ago
8 0

(125 kg)*(9.81 m/s^2)*(10 m)  

W = 12262.5 J

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You make a simple instrument out of two tubes which looks like a flute and extends like a trombone. One tube is placed within an
Marina CMI [18]

Explanation:

We know

the frequency of the tube with one end open and the other end closed follows the given relations as

f_{1} : f_{2} : f_{3} : f_{4} = 1 : 3 : 5 : 7

∴ the 4th allowed wave is f_{4} = 7 f_{1}

                                                                   = \frac{7v}{4l}

We know  f_{4} = 1975 Hz and v = 343 m/s ( as given in question )

∴l = \frac{7\times v}{4\times f_{4}}

 l = \frac{7\times 343}{4\times 1975}

           = 0.303 m

We know that v = f_{4} x λ_{4}

\lambda _{4}= \frac{v}{f_{4}}

\lambda _{4}= \frac{343}{1975}

                            = 0.17 m

Now when the warmer air is flowing, the speed gets doubled and the mean temperature increases. And as a result the wavelength increases but the amplitude and the frequency remains the same.

So we can write

v ∝ λ

or  \frac{v_{1}}{v_{2}}= \frac{\lambda _{1}}{\lambda _{2}}

Therefore, the wavelength becomes doubled = 0.17 x 2

                                                                             = 0.34 m

Now the new length of the air column becomes doubled

∴ l^{'} = 0.3 x 2

                        = 0.6 m

∴ New speed, v^{'} = 2 x 343

                                               = 686 m/s

∴ New frequency is f^{'}=\frac{v^{'}}{4\times l^{'}}

                                 f^{'}=\frac{686}{4\times 0.6}

                                               = 283 Hz

∴ The new frequency remains the same.

Now we know

v_{s} = 12 m/s, v_{o} = 4 m/s, f_{o} = 1975 Hz

Therefore, apparent frequency is f^{'}=f^{o}\left ( \frac{v+v_{s}}{v+v_{o}} \right )

f^{'}=1975\left ( \frac{343+12}{343+4} \right )

            = 2020.5 Hz

7 0
4 years ago
.Which uplift mechanism is pictured here?
zhenek [66]
The correct answer is orographic uplift! Hope this helps
4 0
4 years ago
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Nicolaus Copernicus was considered: a heretic because his theory disagreed with the church a hero because his theory agreed with
PtichkaEL [24]
Nicolaus Copernicus was considered the <span>"Father of Astronomy" because he believed the earth was the center of the solar system. The rest of the choices do not answer the question above.</span>
3 0
4 years ago
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Consider the three displacement vectors
matrenka [14]

Answer:

Explanation:

\overrightarrow{A} = 3\widehat{i}+3\widehat{j}

\overrightarrow{B} = \widehat{i}-4\widehat{j}

\overrightarrow{C} = -2\widehat{i}+5\widehat{j}

(a)

\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}

\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}

\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}

Magnitude of \overrightarrow{D} = \sqrt{2^{2}+4^{2}}

                                                                     = 4.47 m

Let θ be the direction of vector D

tan\theta =\frac{4}{2}

θ = 63.44°

(b)

\overrightarrow{E} = - \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}

\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}

\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}

Magnitude of \overrightarrow{E} = \sqrt{6^{2}+6^{2}}

                                                                     = 8.485 m

Let θ be the direction of vector D

tan\theta =\frac{6}{-6}

θ = 135°

4 0
4 years ago
A baseball has mass 0.145 kg.
Vesnalui [34]

Answer:

a).p=15.67kg*m/s

b). F=7.83N

Explanation:

change in momentum is the subtraction from "after momentum" of the "before momentum"  as momentum is a vector quantity this is a vector subtraction.

initial momentum of ball

m1v1 = (0.145)(44.0) = 6.38 kg-m/s

after momentum

m1v2 = (0.145)(64.0) = 9.28 kg-m/s

since these momentums are 180° opposite one must be called negative so their difference

6.38+9.28= 15.67 kg-m/s

change in momentum = 15.67 kg-m/s ANS a1

Impulse = change in momenutm = 15.67 ANS a2

b)

Impulse =Favg*t

I=(2.00)Favg

Impulse from (a2) = 15.67  

Favg = 15.67/2.00 =

F= 7.83 N

3 0
4 years ago
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