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Keith_Richards [23]
3 years ago
7

A large person and a small person wish to parachute at equal terminal velocities. The larger person will have toa)Jump first fro

m the planeb)Pull upward on the supporting strands to decrease the downward net forcec)Jump lightlyd)Get a larger parachutee)Get a smaller parachute
Chemistry
1 answer:
aliya0001 [1]3 years ago
5 0

Answer:

The larger person will have to get a larger parachute. The answer is D

Explanation:

If the smaller person does nothing, the larger person subsequently accelerate for more time and with larger terminal velocity. In this case the person needs to do something so as to reduce the terminal velocity and can also effectively raise the air resistance. For this reason the person should get a bigger parachute. The air resistance in the opened parachute overwhelms the downward force of the gravity. Whereas the net force as well as the acceleration of the person is upward. The bigger parachute has the ability to grab the greater force. If the parachute, dragging force works in the opposite to that of the force of the gravity, hence the drag force slows the parachute decrease as they fall.

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Which is the greater distance 50 miles or 100 kilometers
slega [8]
The greater distance is 50 miles
4 0
3 years ago
Use standard enthalpies of formation to calculate ΔH∘rxn for the following reaction______.2H2S(g)+3O2(g)→2H2O(l)+2SO2(g) ΔH∘rxn
Ahat [919]

Answer:

-1,103.39KJ/mol

Explanation:

We use the subtract the standard enthalphies of formation of the reactants from that of the products. It must be taken into consideration that the enthalpy of formation of elements and their molecules alone are not taken into consideration. Hence, what we would be considering are the standard enthalpies of formation of H2S, H2O and SO2.

In places where we have more than one mole, we multiply by the number of moles as seen in the balanced chemical equations.

The standard enthalpies of the molecules above are as follows:

H2S = -20.63KJ/mol

H2O = -285.8KJ/mol

SO2 = -296.84KJ/mol

O2 = 0KJ/mol

ΔrH⦵ = [2ΔfH⦵(H2O) + 2 ΔfH⦵(SO2)] − [ΔfH⦵(H2S) + 3

 ΔfH⦵(O2)]

ΔrH⦵ =[(2 × -285.8) + (2 × -296.84)]

-[ 3 × -20.63)]

= (-571.6 - 593.68 + 61.89) = -1,103.39KJ/mol

6 0
4 years ago
A very hot cube of copper metal (32.5 g) is submerged into 105.3 g of water at 15.4 0C and it reach a thermal equilibrium of 17.
zysi [14]

Answer:

The initial temperature of the metal is 84.149 °C.

Explanation:

The heat lost by the metal will be equivalent to the heat gain by the water.  

- (msΔT)metal = (msΔT)water

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-(T final - T initial) = 66.89 °C

T initial = 66.89 °C + T final

T initial = 66.89 °C + 17.3 °C

T initial = 84.149 °C.

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AveGali [126]

Answer:

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Explanation:

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5 0
4 years ago
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Answer:

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