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loris [4]
3 years ago
8

The human ear can respond to an extremely large range of intensities - the quietest sound the ear can hear is smaller than 10-20

times the threshold which causes damage after brief exposure. If you could measure distances over the same range with a single instrument, and the smallest distance you could measure was 1 mm, what would the largest be, in kilometers?
Physics
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

the largest distance we can measure is 10¹⁴ km

Explanation:

Given the data in the question;

Threshold hearing = 10⁻²⁰

smallest distance measured = 1 mm

Largest distance measured will be;

⇒ ( threshold hearing )⁻¹ × smallest distance

= ( 1 / 10⁻²⁰ ) × 1 mm

= 10²⁰ × 1mm

= 10²⁰ mm

we know that; 1000 mm = 10⁶ km

Largest distance = ( 10²⁰ / 10⁶ ) km

= 10¹⁴ km

Therefore, the largest distance we can measure is 10¹⁴ km

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sveticcg [70]

Answer:

a) 72.54°  and  It will take the woman 33.4m

Explanation:

The woman runs with a constant speed V1 = 6m/s

V2 = 20m/s

V1 = V2 cos θ

Cos θ= V1/V2= 6/20

Cos θ= 0.3

Cos^-1 0.3=72.54°

Using Range formular for projectile

R= (V2 Cos θ)/g (V2 Sin θ)^2 +sqrt(V2 Sin θ)^2 + 2gh)

R= (20cos72.54)(2Sin72.54+sqrt(20Sin72.54)^2 + 2×9.8×45

R=33.4m

b )

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Answer:

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3 years ago
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n
satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

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Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

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Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

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= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

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