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3 years ago

What makes it possible for us to see the moon from earth?

Physics

1 answer:

DENIUS [597]3 years ago

3 0

D. Light from the sun is reflected off the moon's surface

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A bullet with a mass of 5.00 x 10^-3 kg is loaded into a gun. The loaded gun has a mass of 0.52 kg. The bullet is fired, causing

beks73 [17]

For the answer to the question above,
220 m/s
mass of the bullet * velocity of the bullet = mass of the gun * velocity of the gun 
0.005 kg * v = 0.52 kg * 2.1 m/s 
-> v = (0.52kg * 2.1 m/s) / 0.005 kg = 218.4 m/s
I hope my answer helped you. Have a nice day!

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4 years ago

a runner makes one lap around a 200m track in 25s, what is the runners (a) average speed and (b) average velocity

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5 0

3 years ago

Which evidence did Ottowa scientists collect in researching PCB concentrations in Alaska?

Sergio [31]

PCBs were found in greater concentrations in lakes where more salmon spawned.

7 0

3 years ago

Read 2 more answers

A jet aircraft is traveling at 260 m/s in horizontal flight. The engine takes in air at a rate of 53.3 kg/s and burns fuel at a

IrinaVladis [17]

Answer:

The thrust of the jet engine is 4188.81 N.

Explanation:

Given that,

Speed = 260 m/s

Rate in air= 53.3 kg/s

Rate of fuel = 3.63 kg/s

Relative speed = 317 m/s

We need to calculate the rate of mass change in the rocket

Using formula of rate of mass

$\dfrac{dM}{dt}=\dfrac{dM_{a}}{dt}+\dfrac{dM_{f}}{dt}$

Put the value into the formula

$\dfrac{dM}{dt}=53.3+3.63$

$\dfrac{dM}{dt}=56.93\ kg/s$

We need to calculate the thrust of the jet engine

Using formula of thrust

$T=\dfrac{dM}{dt}u-\dfrac{dM_{a}}{dt}v$

Put the value into the formula

$T=56.93\times317-53.3\times260$

$T=4188.81\ N$

Hence, The thrust of the jet engine is 4188.81 N.

7 0

4 years ago

An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens

Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

$\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

Putting the values

$\frac{1}{v} +\frac{1}{15} =\frac{1}{14}$

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

$\frac{210}{15} \times \frac{D}{f_e}$

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0

3 years ago