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SSSSS [86.1K]
3 years ago
7

In a choir practice room, two parallel walls are 5.70 m apart. The singers stand against the north wall. The organist faces the

south wall, sitting 0.780 m away from it. To enable her to see the choir, a flat mirror 0.600 m wide is mounted on the south wall, straight in front of her. What width of the north wall can the organist see?
Physics
1 answer:
ololo11 [35]3 years ago
7 0

Answer:

4.98 m

Explanation:

Given that

Width of the mirror, d = 0.6 m

Organist distance to the mirror, s = 0.78 m

Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m

Width of north wall, D?

Using the simple relationship

D/S = d/s, on rearranging

D = dS /s

D = (0.6 * 6.48) / 0.78

D = 3.888 / 0.78

D = 4.98 m

Therefore, we can conclude that the Width of north wall is 4.98 m

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Please Help <br><br> Conservation of energy
Sonja [21]

Answer:

a principle stating that energy cannot be created or destroyed, but can be altered from one form to another.

Explanation:

8 0
3 years ago
While traveling along a highway a driver slows from 32 m/s and comes to a stop with an acceleration of -6 m/s2. How long did it
diamong [38]

Answer: 6s

Explanation:

Vs=32m/s  speed at beginning of slowing down

Vf=0m/s     stop speed

a= -6 m/s²  acceleration

----------------

Use equation for acceleration :

a=(Vf-Vs)/t

a*t=Vf-Vs

t=(Vf-Vs)/a

t=(0-36)/-6

t=-36/-6

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7 0
3 years ago
a rectangular tank measures 12.5 metres long 10.0 wide and 2.0 metre high calculate the mass of the water in the tank when it is
Tom [10]

Answer:

250000kg

Explanation:

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Density=1000kg/m³

Volume=l*w*h

12.5*10.0*2.0=250m³

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7 0
3 years ago
Kelli weighs 440 N, and she is sitting on a playground swing that hangs 0.41 m above the ground. Her mom pulls the swing back an
kifflom [539]

Answer:

Explanation:

Initial height from the ground = .41 m

Final height = 1m

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When she passes through the lowest point , she loses P E

= mgh

= 440 x .59

= 259.6 J

kinetic energy possessed by her

= 1/2 mv²

= .5 x (440/9.8) x 2²

= 89.8 J

Difference of energy is lost due to work by  air friction

work done by  friction = 89.8 - 259.6

= - 169.8 J

4 0
3 years ago
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
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