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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?

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

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4 years ago

.Which uplift mechanism is pictured here?

zhenek [66]

The correct answer is orographic uplift! Hope this helps

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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>

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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