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3 years ago

Mechanical energy is found by____

2 answers:

6 0

An object that possesses mechanical energy is able to do work. In fact, mechanical energy is often defined as the ability to do work. Any object that possesses mechanical energy - whether it is in the form of potential energy or kinetic energy - is able to do work. That is, its mechanical energy enables that object to apply a force to another object in order to cause it to be displaced.

andre [41]3 years ago

3 0

Answer:

I believe the answer is (by ADDING kinetic energy and potential energy)

Are their any answer choices?

Explanation:

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The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box= $u=13m/s$

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

$v^2-u^2=2as$

Substitute the values

$(11.5)^2-(13)^2=2a(8.5)$

$132.25-169=17a$

$-36.75=17a$

$a=\frac{-36.75}{17}=-2.2m/s^2$

We know that

Acceleration= $a=\frac{v-u}{t}$

Substitute the values

$-2.2=\frac{11.5-13}{t}$

$-2.2=\frac{-1.5}{t}$

$t=\frac{1.5}{2.2}=0.68 s$

Hence, the time taken by box to slow by this amount=0.68 s

8 0

3 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Veronika [31]

Answer:

$s=5.79\ km$

$\theta=47^{\circ}$ east of south

Explanation:

Given:

distance of the person form the initial position, $d'=8.4\ km$

direction of the person from the initial position, $47^{\circ}$ north of east

distance supposed to travel form the initial position, $d=5.3\ km$

direction supposed to travel from the initial position, due North

Now refer the schematic for visualization of situation:

$y=d'.\sin47^{\circ}-d$

$y=8.4\times \sin47-5.3$ ...............(1)

$x=d'.\cos47^{\circ}$

$x=8.4\times \cos47^{\circ}$ .................(2)

Now the direction of the desired position with respect to south:

$\tan\theta=\frac{y}{x}$

$\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}$

$\theta=47^{\circ}$ east of south

Now the distance from the current position to the desired position:

$s=\sqrt{x^2+y^2}$

$s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}$

$s=5.79\ km$

4 0

3 years ago

Why are there so few eclipses in 2014?

Sonbull [250]

Answer:

Because the Moon casts a smaller shadow than Earth does, eclipses of the Sun tightly constrain where you can see them. If the Moon completely hides the Sun, even for a moment, the eclipse is considered total.

Explanation:

8 0

3 years ago

Read 2 more answers

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

4 years ago

A wire along the z axis carries a current of 6.8 A in the z direction Find the magnitude and direction of the force exerted on a

Alexeev081 [22]

Answer:

Force is 14.93N along positive y axis.

Explanation:

We know that force 'F' on a current carrying conductor placed in a magnetic field of intensity B is given by

$\overrightarrow{F}=\overrightarrow{Il}\times \overrightarrow{B}$

where L is the length of the conductor

Applying values in the equation we have force F =

$\overrightarrow{F}=6.8\times 6.1\widehat{k}\times 0.36\widehat{i}\\\\\overrightarrow{F}=41.48\widehat{k}\times 0.36\widehat{i}\\\\\therefore \overrightarrow{F}=14.93N\widehat{j}$

Thus force is 14.93N along positive y axis.

3 0

3 years ago