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2 years ago

Hey plz help mesuggest a situations where you might need to use ear defenders

Physics

1 answer:

xenn [34]2 years ago

4 0

You would need to ear defenders (I'm assuming these are ear protectors for use with loud sounds) while firing at a gun range. You could use it for loud construction areas.

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Light passes through a pair of narrow slits with a 0.67-mm separation. It is found that the fourth bright fringe makes an angle

babunello [35]

Answer:

The wavelength of the light is 555 nm.

Explanation:

according to Bragg's law..

n×λ = d×sin(θ)

n is the fringe number

λ is the wavelength of the light

d is the slit separation

θ is the angle the light makes with the normal at the fringe.

7 0

3 years ago

A 65 kg box is lifted by a person pulling a rope a distance of 15 meters straight up at a constant speed. How much Power is requ

Y_Kistochka [10]

Answer:

Power is 1061.67W

Explanation:

Power=force×distance/time

Power=65×9.8×15/9 assuming gravity=9.8m/s²

Power=3185/3=1061.67W

8 0

2 years ago

The acceleration due to gravity on the surface of Jupiter is about 2.5 times the acceleration due to gravity on Earth’s surface.

Ann [662]

Answer: The correct answer is option C.

Explanation:

Weight = Mass × Acceleration

Let the mass of the space probe be m

Acceleration due to gravity on the earth = g

Weight of the space probe on earth = W

$W=m\times g$

Acceleration due to gravity on the Jupiter = g' = 2.5g

Weight of the space probe on earth = W'

$W'=mg'=m\times 2.5g$

$\frac{W'}{W}=\frac{m\times 2.5g}{m\times g}$

$W'=2.5\times W$

The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.

Hence, the correct answer is option C.

6 0

2 years ago

What 2 events increase power output

OverLord2011 [107]

P=I*E
Power (P)
Voltage (E)
Amps (I)

7 0

3 years ago

Read 2 more answers

What fraction of all the electrons in a 25 mg water

mihalych1998 [28]

Answer:

$9.11\times 10^{-15}.$

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.

The charge on one electron, $\rm e=-1.6\times 10^{-19}\ C.$

Let the N number of electrons have charge -40 nC, such that,

$\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.$

Now, mass of one electron = $\rm 9.11\times 10^{-31}\ kg.$

Therefore, mass of N electrons = $\rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.$

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is $25\ \rm mg = 25\times 10^{-6}\ kg.$

Then,

$\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.$

It is the required fraction of mass of the droplet.

3 0

2 years ago