A car traveled 176 miles on 6.4 gallons of fuel the car averaged how many miles per gallon

Which of the following statements are incorrect? a) The state of matter for an object that has a definite volume but not a defin

hoa [83]

Answer:

a), b), d), and e) are correct.

Explanation:

c) is incorrect because an object that has a definite shape cannot be in a gaseous state as gaseous objects take the shape of their container.

a) is correct because objects that are liquid have definite volumes but no definite shapes as they take the shape of their containers,

b) is correct as this is the definition of an object which is in a solid state.

d) is correct as this is the definition of an object which is in a gaseous state.

e) is correct as rigidness, the quality of having a definite shape is only exhibited by objects that are in solid state.

7 0

3 years ago

Suppose 2.10 C of positive charge is distributed evenly throughout a sphere of 1.30-cm radius. 1) What is the charge per unit vo

Anuta_ua [19.1K]

Answer:

$\rho=2.28\times 10^5\ C/m^3$

Explanation:

Given that,

Charge, Q = 2.1 C

The radius of sphere, r = 1.3 cm = 0.013 m

We need to find the charge per unit volume for this situation. It can be calculated a follows:

$\rho=\dfrac{Q}{\dfrac{4}{3}\pi r^3}\\\\\rho=\dfrac{2.1}{\dfrac{4}{3}\pi \times (0.013)^3}\\\\\rho=2.28\times 10^5\ C/m^3$

So, the charge per unit volume is $2.28\times 10^5\ C/m^3$ .

3 0

2 years ago

A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have t

Art [367]

Answer:

3.34×10^-6m

Explanation:

The shear modulus can also be regarded as the rigidity. It is the ratio of shear stress and shear strain

can be expressed as

shear stress/(shear strain)

= (F/A)/(Lo/ . Δx)

Stress=Force/Area

The sheear stress can be expressed below as

F Lo /(A *Δx)

Where A=area of the disk= πd^2/4

F=shearing force force= 600N

Δx= distance

S= shear modulus= 1 x 109 N/m2

Lo= Lenght of the cylinder= 0.700 cm=7×10^-2m

If we make Δx subject of the formula we have

Δx= FLo/(SA)

If we substitute the Area A we have

Δx= FLo/[S(πd^2/4]

Δx=4FLo/(πd^2 *S)

If we input the values we have

(4×600×0.7×10^-2)/10^9 × 3.14 ×(4×10^-2)^2

= 3.35×10^-6m

Therefore, its shear deformation is 3.35×10^-6m

A=area of the disk= πd^2/4

= [3.142×(4×10^-2)^2]/4

7 0

3 years ago

The average intensity of sunlight at the top of the earth's atmosphere in 1390 w/m2. what is the maximum energy that a 34-m x 46

slavikrds [6]

Panel surface area =34m×46m=1,564m^2
total power =1564m^2×1390w/m^2
=2173960watts
now you must calculate total energy.
Energy = power×Time
However time must be in seconds so we multiply 2hrs×60min×60s=7200seconds
7200s×2173960w =15,652,512,000 joules of energy

7 0

3 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