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3 years ago

Which can cause blockage of the ear canal, leading to hearing loss? aging disease earwax noise

Physics

1 answer:

Ivahew [28]3 years ago

5 0

Explanation:

Earwax helps to protect your ears from infection so if you like chewing gum, this helps in removing the wax. But this might not happen at your young age. So if you grow and still likes chewing gum it will remove the wax completely thereby leading to loss on hearing

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If the force of a golf club on a golf ball is 200 N forward, what will the force of the ball on the club be? A. 200 N forward B.

Korolek [52]

The ball should put 200 N of force towards the golfer.

Newton's Third Law is every action has an equal and opposite reaction.

It's the ball exerting 200 N of force towards the club as well, but the opposite reaction is that it flies away.

8 0

4 years ago

Read 2 more answers

What's √ 25what's √ 25

malfutka [58]

Answer:

the square root of 25 is 5

Explanation:

$\sqrt{25}=5$

5 0

3 years ago

g As observed on earth, a certain type of bacteria is known to double in number every 24 hours. Two cultures of these bacteria a

tangare [24]

Answer:

86.4 hrs

Explanation:

The amount of bacteria is initially 1

It doubles every 24 hrs.

After first 24 hrs, the amount = 2

After next 24 hrs = 4

After next 24 hrs = 8

After next 24 hrs = 16

After next 24 hrs = 32

After next 24 hrs = 64

After next 24 hrs = 128

After next 24 hrs = 256

Total time taken to reach 256 = 24 x 8 = 192 hrs

For the bacteria culture on the rocket that travels at a speed of 0.893c relative to the earth, this time is contracted by the relationship

t = t'(1 - ¥^2)^0.5

Where t is the contracted time =?

t' is the time on earth

¥ = v/c

Where v is the speed of the rocket

c is the speed of light

since v = 0.893c

¥ = 0.893

Substituting, we have

t = 192 x (1 - 0.893^2)^0.5

t = 192 x 0.2025^0.5

t = 192 x 0.45 = 86.4 hrs

8 0

3 years ago

The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a

Amanda [17]

Answer:

amount of energy = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = $\frac{excess initial cost}{\Delta M}$

payback period = $\frac{32 + 66}{520.34}$

payback period = 0.188 year

8 0

3 years ago

A stone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9.8 m/s2.) (a) find the dista

Gekata [30.6K]

(a) The stone moves by uniform accelerated motion, with constant acceleration $g=9.81 m/s^2$ directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
$y(t)= y_0 - \frac{1}{2}gt^2= 750 - 4.9 t^2$

(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
$750-4.9 t^2 = 0$
from which we find the time t after which the stone reaches the ground:
$t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s$

(c) The velocity of the stone at time t can be written as
$v(t) = -gt$
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
$v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s$

(d) if the stone has an initial velocity of $v_0 = 6 m/s$ , then its law of motion would be
$y(t)=y_0 - v_0t - \frac{1}{2}gt^2$
and we can find the time it needs to reach the ground by requiring again y(t)=0:
$0=750 - 6t - 4.9 t^2$
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.

5 0

3 years ago