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s2008m [1.1K]
3 years ago
9

What is the minimum diameter mirror on a telescope that would allow you to see details as small as 5.20 km on the moon some 3840

00 km away? Assume an average wavelength of 550 nm for the light received.
cm
Physics
1 answer:
Ksivusya [100]3 years ago
7 0

Answer:

4.96cm

Explanation:

distance  = 5.20km = 5.2 × 10³ m

Wavelength , λ = 550nm = 5.50 × 10⁻⁷m

distance of moon, L = 384000 km = 3.84 × 10⁸m

formula for resolving power of two objects

d = (1.22 × λ ×L) / D

D = (1.22 × λ ×L) / d

    = (1.22 × 5.50 × 10⁻⁷ ×3.84 × 10⁸) / 5.2 × 10³

D = 4.96cm

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What measurements or observations tells you that a car is accelerating
lutik1710 [3]

You need to observe the car at two different times.

-- The first time: 
You write down the car's speed, and the direction it's pointing.

-- The second time:
You write down the car's speed and the direction it's pointing, again.

You take the data back to your lab to analyze it.

-- You compare the first and second speed. If they're different,
then the car had acceleration during the time between the two
observations.

-- You compare the first and second direction.  If those are different,
even if the speeds are the same, then the car had acceleration during
the time between the two observations.

(Remember, "acceleration" doesn't mean "speeding up". 
It means any change in speed or direction of motion.)
8 0
4 years ago
Read 2 more answers
A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0
otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

    F_{1} = 20 N, F_{2} = 25 N, a = -0.9 m/s^{2}

             W = 83 N

         m = \frac{83}{9.81}

             = 8.46

Now, we will balance the forces along the y-component as follows.

       N = W + F_{2}

           = 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

       F_{1} - F_{r} = ma

        20 - F_{r} = 8.46 \times (-0.9)

             F_{r} = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

             F_{r} = \mu \times N

          \mu = \frac{F_{r}}{N}

                    = \frac{7.614}{108}

                    = 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0
4 years ago
Read 2 more answers
A worker pushes horizontally on a large crate with a force of 200 N, and the crate is moved 3.5 m. How much work was done in Jou
sattari [20]

Work = (force) x (distance) =

           (200 N) x (3.5 m)  =  <em>700 joules</em>


6 0
4 years ago
Create a story of a “day in the life of a water drop”. Your story should be creative and include all 6 steps in the water cycle.
pogonyaev

Answer:

Explanation:

Ok so first: the evaporation part, the sun starts to get warmer I ( the water droplet) rises up to the sky to start my evaporation cycle

Condensation: part: when I am in the air I change into a gas and then I change back into a liquid and gather my friends and make a cloud

Precipitation: as it gets to crowded, we can’t hold it anymore, when I cool down I like to sky dive with my cousins, snow, rain, sleet, hail which is called precipitation.

Then finally we land on the ground, we run down hills, and run into lakes surface runoff happens when there’s too many of us so some of us can’t be rain. Infiltration: when some of us soak into the ground cause we can’t make it into the streams and oceans. Ok I can’t help much more cause I’m super busy but if you need more help just message me and I can help thx ! Hope I helped Atleast a bit for you to understand more

5 0
4 years ago
Read 2 more answers
To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 130 m. The car
marysya [2.9K]

Answer:

0.739

Explanation:

If we treat the four tire as single body then

W ( weight of the tyre ) =  mass × acceleration due to gravity (g)

the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r

where v is speed 25.6 m/s and r is the radius of the circle

centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²

net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²

coefficient of static friction between the tires and the road = frictional force / force of normal

frictional force = m × net acceleration / m×g

where force of normal = weight of the body in opposite direction

coefficient of static friction = (7.2567 × m) / (9.81 × m)

coefficient of static friction = 0.739

4 0
3 years ago
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