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3 years ago

9

A standard quantity of a physical property used as a factor to compare other occurring quantities of that property is called a(n

)....
a. rate
b. unit
c. significant figure
d. vector

1 answer:

Brrunno [24]3 years ago

3 0

The correct answer that would complete the given statement above would be option B. UNIT (of measurement). <span>A standard quantity of a physical property used as a factor to compare other occurring quantities of that property is called a unit of measurement. Hope this answers your question. </span>

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What was significant about the clay samples from the Western Hemisphere? How was the Alvarez hypothesis modified in response?

Elenna [48]

Answer:

Either Answer you Put is fine i put one as an answer and the other is the sample response and got it right.

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Sample Response:

Samples from the Western Hemisphere contained significantly higher amounts of shock-fractured quartz. This led Walter and Luis Alvarez to hypothesize that the asteroid impact site was in the Western Hemisphere.

Explanation:

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4 years ago

What type of device is a refrigerator?

Tamiku [17]

A heat pump that uses work to move heat

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3 years ago

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s^2 for 20s.

svet-max [94.6K]

Answer:

A) The distance between the stations is 1430m

B) The time it takes the train to go between the stations is 80s

Explanation:

First we will calculate the distance covered for the first 20s.

From one the equations of kinematics for linear motion

$S = ut + \frac{1}{2}at^{2} \\$

Where $S$ is distance traveled

$u$ is the initial velocity

$t$ is time

and $a$ is acceleration

Since the train starts from rest, $u$ = 0 m/s

Hence, for the first 20s

$a =$ 1.1 m/s²; $t =$ 20s, $u$ = 0 m/s

∴ $S = ut + \frac{1}{2}at^{2} \\$ gives

$S = (0)(20) + \frac{1}{2}(1.1)(20)^{2}$

$S = \frac{1}{2}(1.1)(20)^{2}$

$S =$ 220m

This is the distance covered in the first 20s.

The train then proceeds at constant speed for 1100m.

Now, we will calculate the speed attained here

From

$v = u +at$

Where $v$ is the final velocity

Hence,

$v = 0 + 1.1(20)$

$v = 1.1(20)$

$v =$ 22 m/s

This is the constant speed attained when it proceeds for 1100m

The train then slows down at a rate of 2.2 m/s² until it stops

We can calculate the distance covered while slowing down from

$v^{2} = u^{2} + 2as$

The initial velocity, $u$ here will be the final velocity before it started slowing down

∴ $u$ = 22 m/s

The final velocity will be 0, since it came to a stop.

∴ $v$ = 0 m/s

$a = -$ 2.2 m/s² ( - indicates deceleration)

Hence,

$v^{2} = u^{2} + 2as$ gives

$0^{2} =22^{2} +2(-2.2)s$

$0=22^{2} - (4.4)s\\4.4s = 484\\s = \frac{484}{4.4} \\s = 110m$

This is the distance traveled while slowing down.

A) The distance between the stations is

220m + 1100m + 110m

= 1430m

Hence, the distance between the stations is 1430m

B) The time it takes the train to go between the stations

The time spent while accelerating at 1.1 m/s² is 20s

We will calculate the time spent when it proceeds at a constant speed of 22 m/s for 1100m,

From,

$Speed =\frac{Distance}{Time}\\$

Then,

$Time = \frac{Distance}{Speed}$

$Time = \frac{1100}{22}$

Time = 50 s

And then, the time spent while decelerating (that is, while slowing down)

From,

$v = u + at\\0 = 22 +(-2.2)t\\2.2t = 22\\t = \frac{22}{2.2} \\t= 10 s$

This is the time spent while slowing down until it stops at the station.

Hence, The time it takes the train to go between the stations is

20s + 50s + 10s = 80s

The time it takes the train to go between the stations is 80s

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3 years ago

A thin film of alcohol (n = 1.36) lies on a flat glass plate (n = 1.51). When monochromatic light, whose wavelength λ can be cha

Shkiper50 [21]

Answer:

The thickness of the film, t = 470.59 nm

Explanation:

$\lambda_{film} = \frac{\lambda}{n_{film} }$

For constructive interference, the net phase change, $\phi = \phi_{2} - \phi_{1} = 2\pi m_{1}$

Then the thickness, $t = 0.5 \lambda_{film} m_{1}$

m₁ = 1,2,3........

$t = 0.5\frac{\lambda_{1} }{n_{film} } m_{1}$ ..............(i)

For destructive interference, the net phase change is $\phi = \phi_{2} - \phi_{1} = 2(m_{2} + 1)\pi$

m₂ = 1,2,3........

thickness, $t = 0.5\frac{\lambda_{1} }{n_{film} }(2 m_{2}+1)$ ..............(ii)

Equating (i) and (ii)

$0.5\frac{\lambda_{1} }{n_{film} } m_{1} = 0.5\frac{\lambda_{1} }{n_{film} }(2 m_{2}+1)$

$\frac{2m_{2} + 1}{2m_{1} } = \frac{\lambda_{1} }{\lambda_{2}} = \frac{640}{512}$

$\frac{2m_{2} + 1}{2m_{1} } = 1.25$ .............(iii)

From (iii), When m₁ =2, m₂ = 2

The thickness , $t = 0.5\frac{\lambda_{1} }{n_{film} } m_{1}$ then becomes

$t = 0.5\frac{640}{1.36} *2$

t = 470.59 nm

7 0

3 years ago

A 1500-kg car doubles its speed from 50 km/h to 100 km/h. By how many times does the kinetic energy from the car's forward motio

Marat540 [252]

Answer:

The kinetic energy increases 4 times

Explanation:

Given,

The mass of the car, m = 1500 Kg

The speed of the car doubles from 50 Km/h to 100 Km/h

The kinetic energy at 50 Km/h

K.E = ½mv²

= ½ x 1500 x 50²

= 1875000 J

The kinetic energy of the car at 100 Km/h

K.E = ½mv²

= ½ x 1500 x 100²

= 7500000 J

Hence, the final K.E is 4 times the initial K.E

6 0

3 years ago