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blondinia [14]
3 years ago
14

What are 3 ways objects can be grouped based on how much light they let shine/pass through them.

Physics
1 answer:
Wittaler [7]3 years ago
5 0

Answer:

Transparent = lets light pass through without significant interference

Translucent = lets light pass through with significant interference

Opaque = Doesn't let light pass, instead absorbs it

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The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
2 years ago
A boy jogs 250 meters in 110s what is the average speed
Artyom0805 [142]
The rule to get the average speed is as follows:
average speed = average distance / average time

We are given that:
distance = 250 m
time = 110 sec

Substitute with the givens in the above equation to get the average speed as follows:
average speed = 250/110 = 25/11 meters/sec
3 0
3 years ago
A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an
Vlada [557]

Answer:

Explanation:

Given

volume V=16 ft^3

Suppose base is square with side L

height of crate is h

Volume V=L^2\times h

16=L^2\times h

Cost of top and bottom area c_1=3L^2

Cost of Side area c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}

Total Cost C=c_1+c_2

Total Cost C=3L^2+\frac{128}{L}

Differentiate C w.r.t Length

\frac{dC}{dL}=6L-\frac{128}{L^2}

L^3=\frac{128}{6}

L=2.75 ft

h=\frac{16}{2.75^2}=11.46 ft

Dimensions are L\times L\times h=2.75\times 2.75\times 11.46    

6 0
3 years ago
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