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Alex17521 [72]
3 years ago
5

If you wanted to detect x rays coming from the sun, where would you place the detector? Why?

Physics
2 answers:
Damm [24]3 years ago
4 0
You would have to place your sensor above earth's atmosphere because it blocks out nearly all x-rays. this is why we have the Chandra observatory

hope this helps
Nady [450]3 years ago
3 0
If you wanted to detect X-rays coming from the sun, the detector would have to be placed in orbit around Earth since most X-rays are absorbed the atmosphere.

Hope this helped :)
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A nanometer is a unit of mass, whereas a nanosecond is a unit of time. Question 1 options: True False
Alenkasestr [34]

Answer:

True.

Explanation:

A nanometer is a unit of mass, whereas a nanosecond is a unit of time. To convert 1.3 hours to minute, you would multiply by 1 h / 60 min. Kilometer is a unit of length, whereas kilogram is a unit of mass. True.

6 0
2 years ago
Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is foc
vekshin1

Answer:

Explanation:

Part A) Using

light intensity I= P/A

A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B)  Using

I=c*ε*E^2/2

rearrange to solve for E= \sqrt{((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

I= the already solved light intensity= 8.85*10^10 W/m^2

amplitude of the electric field E= \sqrt{(9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= \sqrt{(1.8*10^11) / (2.66*10^-3) = \sqrt{(6.8*10^13) = 8.25*10^6 V/m    

 

8 0
3 years ago
An electric field of 3x10^16 n/c is needed to create a spark in the air. If the charged particles in the field are separated by
meriva
The answer of this problem is positive 3
6 0
3 years ago
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0
3 years ago
What charges are needed in the objects to attract both objects?
kvasek [131]

Answer:

both

Explanation:

8 0
3 years ago
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