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Kamila [148]
3 years ago
8

Please somebody answer I desperately need help!

Physics
2 answers:
padilas [110]3 years ago
8 0
Preserved fossil<span> (like a fossil in amber, ice or tar.</span>
vredina [299]3 years ago
4 0
Amber. The insects would get stuck in the Amber, which is really thick, and it would form around the insect.
I hope this helps.
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A carpenter on the roof of a building accidentally drops herhammer. As the hammer falls it passes two windows of equal height,a)
nadezda [96]

Answer:

Explanation:

a) the speed increment of the hammer as it drops past the first window, is greater than that of the speed of the hammer as it drops past the second window. This can also be translated as saying that the hammer spent more time at the second window.

b) III

The best answer would be answer III, The hammer spends more time dropping past window 1, which I had already included in my explanation in (a) above.

4 0
3 years ago
The amplitude of a paricular wave is 4.0 m. The crest to trough distance
kozerog [31]

Answer:

The crest to trough distance = 8 m

Explanation:

Given that,

The amplitude of a particular wave is 4.0 m.

We need to find the crest to trough distance.

We know that,

Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.

It means,

Distance from crest to trough = 2(Amplitude)

= 2(4)

= 8 m

Hence, the crest to trough distance is equal to 8 m.

6 0
2 years ago
Look at the graph pic and answer the question correctly!
SCORPION-xisa [38]

Answer:

B)

Explanation:

That the time period of which they stop.

4 0
3 years ago
Read 2 more answers
Does displacement = Δx?
Volgvan
Typically no. Displacement can be in multiple directions as a vector. of something is traveling only along x, then it would be true though this is usually not the case.
8 0
3 years ago
Read 2 more answers
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
3 years ago
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