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Soloha48 [4]
2 years ago
6

Professional application dr. john paul stapp was u.s. air force officer who studied the effects of extreme deceleration on the h

uman body. on december 10, 1954, stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! calculate his (a) acceleration and (b) deceleration. express each in multiples of ? (9.80 m/s2) by taking its ratio to the acceleration of gravity.
Physics
1 answer:
aleksley [76]2 years ago
5 0
A) Acceleration= (282m/s-0m/s)/5.0s= 56.4m/s^2
b) Deceleration=(0m/s-282m/s)/1.40s=201.4m/s^2 (here we reject negative because deceleration already take into account negative)
c) for the first one the multiple will be 5.76g
for the second one it will be 20.6g
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Answer:

the work done by the 30N force is 4156.92 J.

For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:

W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J

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two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t
Tomtit [17]

The charges have opposite sign and magnitude 6 \mu C

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

q_1 = q_2 = q

So we can rewrite the equation as

F=\frac{kq^2}{r^2}

And solving for q:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

q_1 = 6 \mu C\\q_2 = -6 \mu C

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

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2 years ago
HELP PLS!<br> (LOOK AT THE PICTURE)
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The answer is b !!!! Hope it helps
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2 years ago
xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi
slega [8]

Answer:

Approximately \rm 90\; kJ.

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction (E^{\circ}(\text{cell})) is equal to

E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode}).

There are two half-reactions in this question. \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu and \rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb. Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of E^{\circ}(\text{cell}) should be positive.

In this case, E^{\circ}(\text{cell}) is positive only if \rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu is the reaction takes place at the cathode. The net reaction would be

\rm Cu^{2+} + Pb \to Cu + Pb^{2+}.

Its cell potential would be equal to 0.339 - (-0.130) = \rm 0.469\; V.

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell}),

where

  • n is the number moles of electrons transferred for each mole of the reaction. In this case the value of n is 2 as in the half-reactions.
  • F is Faraday's Constant (approximately 96485.33212\; \rm C \cdot mol^{-1}.)

\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}.

5 0
2 years ago
Help<br><br><br> What are energy and work
mylen [45]

Answer:

<em>In physics, energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object. Energy is a conserved quantity; the law of conservation of energy states that energy can be converted in form, but not created or destroyed.</em>

<em />

<em>In physics, energy is the quantitative property that must be transferred to an object in order to perform work on, or to heat, the object. Energy is a conserved quantity; the law of conservation of energy states that energy can be converted in form, but not created or destroyed.</em>

Explanation:

<h2><u><em>HOPE THIS HELPS</em></u></h2>
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