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4 years ago

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Anya recorded the temperatures of four different smooth materials after they were placed under a heat lamp for thirty minutes. A

2-column table with 4 rows. The first column labeled Material has entries W, X, Y, Z. The second column labeled Temperature (degrees Fahrenheit) has entries 87, 99, 75, 105. Which conclusion is best supported by the information in the chart? Material W is less absorbent than material Z. Material Y is the least reflective. Material X is more reflective than material W. Material Z is the least absorbent.

2 answers:

bekas [8.4K]4 years ago

4 0

Answer:

B

Explanation:

Llana [10]4 years ago

3 0

Answer:

Material Y is the least reflective.

Explanation:

B

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A cylinder with a piston contains 0.200 mol of nitrogen at 1.50×105 Pa and 320 K . The nitrogen may be treated as an ideal gas.

Alja [10]

Answer:

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

Explanation:

First gas is compressed isobarically such that its volume is half of initial volume

So its temperature is also half

So heat given in this process is given as

$Q = nC_p \Delta T$

for diatomic gas we have

$C_p = \frac{7}{2} R$

so we will have

$Q = 0.200(\frac{7}{2}R)(160 - 320)$

$Q = -930.7 J$

Now in adiabatic process heat is not transferred

so in this process

$Q = 0$

so we have

$T_1V_1^{1.4-1} = T_2V_2^{1.4-1}$

$(160)(\frac{V}{2})^{0.4} = T_2(V)^{0.4}$

$T_2 = 121.26 K$

Now it is again reached to original pressure

so temperature will become initial temperature

so heat given in that part

$Q_3 = nC_v\Delta T$

here we know that

$C_v = \frac{5}{2}R$

$Q_3 = (0.200)(\frac{5}{2}R)(320 - 121.26)$

$Q_3 = 825.76 J$

So total heat given to the system is

$Q = -930.7 + 0 + 825.76$

$Q = -105 J$

Also we know that for cyclic process change in internal energy is always ZERO

3 0

3 years ago

13. Litmus paper can be used to determine if a solution is an acid or a base. There are two types of

Over [174]

Answer:

The main use of litmus is to test whether a solution is acidic or basic. Light Blue litmus paper turns red under acidic conditions and red litmus paper turns blue under basic or alkaline conditions, with the color change occurring over the pH range 4.5–8.3 at 25 °C (77 °F). Neutral litmus paper is purple.

Explanation:

7 0

3 years ago

1. Determine the energy released per kilogram of fuel used.

yarga [219]

200 MeV of energy
E1/E2=7.61=8

U is equal to 1 kilogram or 1000 g.
There are 6.02310 23 atoms in one mole, or 235 g, of uranium. Therefore, 6.02310 23 atoms are present in 1000 g of 92/235 U.
It is understood that one atom releases 200 MeV of energy during its fission.

As a result, the energy released from the fission of one kilogram of 92/235 is given by E 2 = 6.02310 23 1000200/235 =5.10610 26 MeV E1/E2=7.61=8

In light of this, the energy released during the fusion of one kilogram of hydrogen is roughly eight times greater than the energy generated during the fission of one kilogram of uranium.

To learn more about Fission please visit -
brainly.com/question/27923750
#SPJ1

6 0

2 years ago

(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface

navik [9.2K]

Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

3 0

3 years ago

A 80 g particle is moving to the left at 22 m/s . How much net work must be done on the particle to cause it to move to the righ

AysviL [449]

Answer:

W= 38.4 J

Explanation:

Given that

m = 80 g= 0.08 kg

Initial speed ,u= 22 m/s

Final speed ,v= 38 m/s

The change in the kinetic energy of the particle

$\Delta KE=\dfrac{1}{2}m(v^2-u^2)$

$\Delta KE=\dfrac{1}{2}\times 0.08\times (38^2-22^2)\ J$

ΔKE= 38.4 J

We know that

Work done by all the forces =Change in the kinetic energy

That is why net work done = 38.4 J.

W= 38.4 J

Therefore the answer will be 38.4 J.

8 0

3 years ago