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2 years ago

What could damage the eardrum and bones of the middle ear?

Physics

1 answer:

klio [65]2 years ago

8 0

Answer:

B or D

Explanation:

I wanna say B but I don't really know if it's a trick question

my explanation for D is that anvils are usually 110 decibles and ear drums usually are damaged when exposed to 82dbs or higher at a consistant rate so I'd imagine that'd be the answer I hoped this helped

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You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity

Anarel [89]

The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).

The force exerted on the book by gravity has magnitude

F = mg = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N

You raise the book 1.0 m in the opposite direction, so the work done is

W = (10 N) (-1.0 m) = -10 J

5 0

3 years ago

Air "breaks down" when the electric field strength reaches 3 × 106 n/c, causing a spark. A parallel-plate capacitor is made from

Lera25 [3.4K]

Electric field between the plates of parallel plate capacitor is given as

$E = \frac{Q}{A\epsilon_0}$

here area of plates of capacitor is given as

$A = 0.055 * 0.055$

$A = 3.025 * 10^{-3}$

also the maximum field strength is given as

$E = 3 * 10^6 N/C$

now we will plug in all data to find the maximum possible charge on capacitor plates

$3 * 10^6 = \frac{Q}{3.025 * 10^{-3}*8.85 * 10^{-12}}$

$Q = 8.03 * 10^{-8} C$

so the maximum charge that plate will hold will be given by above

3 0

2 years ago

Read 2 more answers

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle

tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

$-h= vsinA\times t-gt^2/2$

putting values h=15 m, v=0.8

$-15 = 0.8vt - 4.9t^2$ ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

$-15 = 0.8\times40/0.6 - 4.9t^2$

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = $-H =v sinA t - gt^2/2$

⇒ $4.9t^2 - 8.9\times0.8t - 100 = 0$

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0

2 years ago

An object is pushed with a force of 400 N for 20 meters; what is the work done on the object? Solve using W=F*d

nika2105 [10]

Answer:

W = Fd = 400(20) = 8000 J

Explanation:

5 0

2 years ago

Help please physics!!!

Hitman42 [59]

Answer:

I think c is the answer.....

6 0

2 years ago

Read 2 more answers