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irga5000 [103]
3 years ago
6

20. Consider a model steel bridge that is 1/100 the exact scale of the real bridge that is to be built. a. If the model bridge w

eighs 50 N, what will the real bridge weigh? b. If the model bridge does not appear to sag under its own weight, is this evidence the real bridge, built exactly to scale, will not appear to sag either?
Physics
1 answer:
Veseljchak [2.6K]3 years ago
6 0
The model bridge captures all the structural attributes of the real bridge, at a reduced scale.

Part a.
Note that volume is proportional to the cube of length. Therefore the actual bridge will have 100^3 = 10^6 times the mass of the model bridge.

Because the model bridge weighs 50 N, the real bridge weighs
(50 N)*10^6 = 50 MN.

Part b.
The model bridge matches the structural characteristics of the actual bridge.
Therefore the real bridge will not sag either.
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Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite r_{1}= 8.20\times10^{7}

Orbital radius of second satellite r_{2}=7.00\times10^{7}\ m

Mass of first satellite m_{1}=53.0\ kg

Mass of second satellite m_{2}=54.0\ kg

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

v=\sqrt{\dfrac{GM}{r}}

From this relation,

v_{1}\propto\dfrac{1}{\sqrt{r}}

Now, \dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}

v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}

Put the value into the formula

v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}

v_{2}=5195.16\ m/s

Hence, The orbital speed of this second satellite is 5195.16 m/s.

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