Motion maps for two objects, Y and Z, are shown.

An object is 6.0 cm in front of a converging lens with a focal length of 10 cm.Use ray tracing to determine the location of the

Masja [62]

As we know that

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

here we know that

$d_0 = 6 cm$

$f = 10 cm$

now from above equation we have

$\frac{1}{d_i} + \frac{1}{6} = \frac{1}{10}$

$d_i = -15 cm$

so image will form on left side of lens at a distance of 15 cm

This image will be magnified and virtual image

Ray diagram is attached below here

8 0

2 years ago

Eratosthenes determined the circumference of Earth by conducting an experiment. Put his steps in order as they correlate to the

galben [10]

Solution: The correct order is: C, A, B

The statement of the problem:

How can we prove Earth is round and calculate its circumference?

Hypotheis:

If the sun casts shadows at different angles at the same time of day in different places, we can determine how much Earth curves.

If the Earth was flat, the angle measured at different places at the same time of the day would be same.

Observation:

In Syene, the sun's rays are vertical at noon. At the same time in Alexandria, the rays are 7.2 degrees from the vertical.

5 0

2 years ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Burka [1]

Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .

At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .

At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .

7 0

2 years ago

Read 2 more answers

A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga

andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

Cp-Cv= R

Cp=7/2 R

Now by putting the values

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

$\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}$

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

Process is adiabatic

$\Delta S_{surr}=0$

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

$\Delta S_{univ}= 20.79 +0$

$\Delta S_{univ}= 20.79$

3 0

2 years ago

Determine the specific volume of refrigerant-134a at 1 MPa and 50°C, using (a) the ideal-gas equation of state and (b) the gener

Andrej [43]

Answer:

( a ) The specific volume by ideal gas equation = 0.02632 $\frac{m^{3} }{kg}$

% Error = 20.75 %

(b) The value of specific volume From the generalized compressibility chart = 0.0142 $\frac{m^{3} }{kg}$

% Error = - 34.85 %

Explanation:

Pressure = 1 M pa

Temperature = 50 °c = 323 K

Gas constant ( R ) for refrigerant = 81.49 $\frac{J}{kg k}$

(a). From ideal gas equation P V = m R T ---------- (1)

⇒ $\frac{V}{m}$ = $\frac{R T}{P}$

⇒ Here $\frac{V}{m}$ = Specific volume = v

⇒ v = $\frac{R T}{P}$

Put all the values in the above formula we get

⇒ v = $\frac{323}{10^{6} }$ ×81.49

⇒ v = 0.02632 $\frac{m^{3} }{kg}$

This is the specific volume by ideal gas equation.

Actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.02632 - 0.021796 = 0.004524 $\frac{m^{3} }{kg}$

% Error = $\frac{0.004524}{0.021796}$ × 100

% Error = 20.75 %

(b). From the generalized compressibility chart the value of specific volume

$\frac{V}{m}$ = v = 0.0142 $\frac{m^{3} }{kg}$

The actual value = 0.021796 $\frac{m^{3} }{kg}$

Error = 0.0142 - 0.021796 = $\frac{m^{3} }{kg}$

% Error = $\frac{- 0.0076}{0.021796}$ × 100

% Error = - 34.85 %

3 0

2 years ago