Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona

Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

$k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m$

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

$E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J$

Therefore, the additional work needed is

$\Delta E=E'-E=810-90=720 J$

Learn more about work:

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7 0

3 years ago

nert xenon actually forms many compounds, especially with highly electronegative fluorine. The ΔH o f values for xenon difluorid

loris [4]

Answer:

For Xenon fluoride, the average bond energy is 132kj/mol

For tetraflouride,the average bond energy is 150.5kj/mol.

For hexaflouride, the average bond energy is 146.5 kj/mol

Explanation:

For xenon fluoride

105/2 = 52.5

For F-F

159/2 = 79.5

Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole

For tetraflouride

284/4 = 71

For F-F

159/2 = 79.5

Average bond energy = 79.5 + 71 = 150.5kj/mol

For hexaflouride

402/6 = 67

F-F = 159/2 = 79.5

Average bond energy = 67 + 79.5 = 146.5kj/ mol

3 0

3 years ago

when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of

Ahat [919]

<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q</h2>

Explanation:

Specific heat capacity

It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .

It is given as :

Heat absorbed = mass of substance x specific heat capacity x rise in temperature

or ,

Q= m x c x t

In above question , it is given :

For Q

mass of Q = m

Temperature changed =T₂/2

Heat supplied = x

Q= mc t

or

X=m x C₁ X T₁

or, X =m x C₁ x T₂/2

or, C₁=X x 2 /m x T₂ (equation 1 )

For another quantity : P

mass of P =m/2

Temperature= T₂

Heat supplied is same that is : X

so, X= m/2 x C₂ x T₂

or, C₂=2X/m. T₂ (equation 2 )

Now taking ratio of C₂ to c₁, We have

C₂/C₁= 2X /m.T₂ /2X /m.T₂

so, C₂/C₁= 1/1

so, the ratio is 1: 1

8 0

3 years ago

You have a 78.7 mF capacitor initially charged to a potential difference of 11.5 V. You discharge the capacitor through a 3.03 Ω

Aloiza [94]

Answer:

$\tau=0.23\;second$

Explanation:

Given,

$C=78.7\;mF\\V=11.5\;V\\R=3.03\;\Omega\\$

Time constant

$\tau=RC\\\tau=78.7\times10^{-3}\times3.03\\\tau=238.461\times10^{-3}\;second\\\tau=0.23\;second$

8 0

3 years ago

Explain the increase in pressure of a gas when its volume is decreased at constant<br> temperature.

Dmitriy789 [7]

Answer:

For a gas held at constant temperature, we can apply Boyle's law, which states that the product between the gas pressure and its volume is constant:

$PV=const.$

where

P is the pressure

V is the volume

As we see from the equation, P and V are inversely proportional to each other: this means that when the volume is decreased, the pressure increases, and vice-versa. The reason for that is that when the volume is decreased, the gas is compressed, so the molecules of the gas come closer to each other, so they collide more frequently with the wall of the container, exerting therefore a greater pressure.

7 0

3 years ago