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2 years ago

Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?

Physics

1 answer:

zhuklara [117]2 years ago

4 0

Answer:

It will not hit.

Explanation:

In this problem, we need to tell that was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s.

When it will hit the target, final velocity, v = 0

Using third equation of motion as follows :

$v^2-u^2=2as$

Here, a =-g

$s=\dfrac{u^2}{2g}\\\\s=\dfrac{(20)^2}{2\times 9.8}\\\\=20.40\ m$

The target is in a straight line 50m away on the x-axis. As 50m is far from 20.40 m and that’s why it won’t hit.

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A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

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Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

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So:

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And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

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• Heterotopic protons are those that when substituted by the same substituent, are structurally different. They are not similar, diastereotopic or enantiotopic.

• Diastreotopic protons refers to two protons in a molecule which, if replaced by the same substituent, would generate compounds that are diastereomers. Diastereotopic groups are often, but not always, identical groups attached to the same atom in a molecule containing at least one chiral center.

For example, the two hydrogen atoms of the C3 carbon in (S)-2-bromobutane are diastereotopic (shown in the attached image). Replacement of one hydrogen atom with a bromine atom will produce (2S,3R)-2,3-dibromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the diastereomer (2S,3S)-2,3-dibromobutane.

• Homotopic protons in a compound are equivalent protons. Two protons A and B are homotopic if the molecule remains the same (including stereochemically) when the protons are interchanged with some other atom (substituent) while the remaining parts of the molecule stay fixed. Homotopic atoms are always identical, in any environment.

For example, ethane, the two H atoms on C1 and C2 carbons on the same side (as shown in the attached image) are homotopic as they exhibit the phenomenon described above.

• Enantiotopic protons are two protons in a molecule which, if one or the other were replaced (by the same substituent), would generate a chiral compound. The two possible compounds resulting from that replacement would be enantiomers.

For example, in the attached image to this answer, the two hydrogen atoms attached to the second carbon in butane are enantiotopic. Replacement of one hydrogen atom with a bromine atom will produce (R)-2-bromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the enantiomer (S)-2-bromobutane.

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