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2 years ago

10

Dao makes a table to identify the variables used in the equations for centripetal acceleration. A 2 column 5 rows. The first col

umn is labeled Variable with entries a Subscript c Baseline, T, r, v. The second column is labeled Quantity with entries blank, X, blank, Y. What quantities belong in cells X and Y? X: centripetal acceleration Y: period X: tangential speed Y: radius X: radius Y: centripetal acceleration X: period Y: tangential speed

2 answers:

Viefleur [7K]2 years ago

8 0

Answer:

Column X: Tangential Speed

Column Y: Radius

Explanation:

Zanzabum2 years ago

5 0

Answer:

Column X. Tangential Speed

Column Y. radius

Explanation:

The equation for centripetal acceleration is

$a_{c}$ = v² / r

Where v is the tangential velocity of the body and the radius of curvature.

To analyze this equation you must place the tangential velocity in one column and in the other the turning radius

Let's check the answers

Column X. Tangential Speed

Column Y. radius

This is the correct answer.

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Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi

Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

$\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}$

here we know that

$\mu_1 = 1.473$

$h_o = 5 cm$

$\mu_2 = 1$

now by above formula

$\frac{1.473}{5} = \frac{1}{h_i}$

$h_i = 3.39 cm$

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

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2 years ago

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

$\Delta S_{surr}=0$

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

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

$\Delta S_{univ}= 20.79$

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2 years ago

Write the meaning of an object has 2 meter length

Ivahew [28]

Answer:

The object has 2 meter length. This means the length is any quantity with a dimension distance. The definition of the length is how long something is or amount of space. In the given data it is stated that, the object is something that has a length of 2 meter.

<em>Let's take examples to understand. </em>

For example a thread or a table is an object which has a total length of 2 meters.

Another example is something we are measuring it gives us a result of 2 meters of length by using a meter scale or meter tape.

Length is a measure of distance and it is a fundamental quantity. Meter is a international system of units (SI units).

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2 years ago

The air surrounding an airplane in flight exerts a drag force that acts opposite to the airplane's motion. When an Airbus A380 i

Rainbow [258]

Answer:

$4.32\cdot 10^5 hp$ , $3.22\cdot 10^8 W$

Explanation:

The jet is flying at constant velocity: this means that its acceleration is zero, so the net force acting on the jet is also zero.

Therefore, we can write:

$4F_T - F_d = 0$

where

$F_T = 322,000 N$ is the thrust force generated by each engine of the jet

$F_d$ is the drag force

Solving for Fd,

$F_d = 4 F_T = 4(322,000)=1.288\cdot 10^6 N$

The velocity of the jet is

$v=250 m/s$

So, the rate at which the drag force does work (which is the power) is

$P=F_d v$

and substituting

$F_d = 1.288\cdot 10^6 N\\v = 250 m/s$

we find

$P=(1.288\cdot 10^6)(250)=3.22\cdot 10^8 W$

Converting into horsepower,

$P=\frac{3.22\cdot 10^8}{746}=4.32\cdot 10^5 hp$

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2 years ago

You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you

lord [1]

Answer:

The speed of light will be c=3x10^8m/s

Explanation:

This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c

5 0

2 years ago