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3 years ago

12

Calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30 meter per seco

nds?

2 answers:

mamaluj [8]3 years ago

6 0

Explanation:

u = 0m/s

v = 30m/s

a = g = 10m/s²

s = H = ?

using the formula,

v² = u² + 2as

30² = 0² + 2(10)(H)

900 = 20H

H = 900/20

H = 45m

andre [41]3 years ago

5 0

Answer:

the answer is

45m - height

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A current I flows down a wire of radius a.

Helga [31]

Answer:

(a) $K = \frac{I}{2\pi a}$

(b) $J = \frac{I}{2\pi as}$

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

$K = \frac{dI}{dL}$

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

$dL = 2\pi r$

The radius of the wire = a

$dL = 2\pi a$

The surface current density $K = \frac{I}{2\pi a}$

(b) The current density is inversely proportional

$J \alpha s^{-1}$

$J = \frac{k}{s}$ ......(1)

k is the constant of proportionality

$I = \int\limits {J} \, dS$

$I = J \int\limits \, dS$ ........(2)

substituting (1) into (2)

$I = \frac{k}{s} \int\limits\, dS$

$I = k \int\limits^a_0 \frac{1}{s} {s} \, dS$

$I = 2\pi k\int\limits\, dS$

$I = 2\pi ka$

$k = \frac{I}{2\pi a}$

substitute $J = \frac{k}{s}$

$J = \frac{I}{2\pi as}$

7 0

3 years ago

A car slows down from 27.7 m/s <br> to 10.9 m/s in 2.37 s. <br> What is its acceleration?

Sunny_sXe [5.5K]

Answer:

- 7.088 m/s²

Explanation:

As we know that,

★ Acceleration = Change in velocity/Time

→ a = (v - u)/t

Here,

Initial velocity (u) = 27.7 m/s
Final velocity (v) = 10.9 m/s

→ a = (10.9 m/s - 27.7 m/s)/2.37 s

→ a = -16.8/2.37 m/s²

→ <u>a</u><u> </u><u>=</u><u> </u><u>-</u><u>7</u><u>.</u><u>0</u><u>8</u><u>8</u><u> </u><u>m/s²</u> [Answer]

Negative sign denotes that the velocity is decreasing.

4 0

3 years ago

Read 2 more answers

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 $96485.33212\; \rm C \cdot mol^{-1}$ .)

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

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3 years ago

Describe two forms of energy-efficient transportation.

vodka [1.7K]

Electricity which may efficient transportation if you develop your way depending contact metal basis , such as electric train .

-Water is from of heat-energy using solar power which run a turbine may be wheel on a railway path.

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3 years ago

Another athlete using a different spring exerts an average force of 400 N to enable her

babymother [125]

Answer:84Nm

Explanation:

force=400N

Distance=0.210m

Workdone=force x distance

Workdone=400 x 0.210

Workdone=84Nm

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3 years ago