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3 years ago

What is the value of the fixed points of a temperature scale

Physics

1 answer:

AleksAgata [21]3 years ago

8 0

Answer:

To calibrate a thermometer is to mark a thermometer so that you can use it to measure temperature accurately. A fixed point is a standard degree of hotness or coldness such as the melting point of ice or boiling point of water.

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A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago

Which statement is true? Which statement is true? An orbital that penetrates into the region occupied by core electrons is more

masya89 [10]

Hey there!:

Here the Statement - D is correct.

Because Orbitals containing the core electrons are more attracted towards nuclear charge and hence less shilded from nuclear charge than an orbital that doesn't penetrate. Also due to more attraction between the orbital containing core electron and nucleus, it will have less energy.

Hope this helps!

6 0

3 years ago

A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.

Luba_88 [7]

Answer:

The final equilibrium temperature of the system is $T = 12.48^oC$

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice $M_{i}$ = 129 g = 0.129 kg

Mass of the steam $M_s$ = 19 g = 0.019 kg

Initial temperature is $T_i$ = 0°C

Temperature of steam $T_s$ = 100°C

Following the change of state of water in the question

The energy required by ice to change to water is mathematically given as

$Q_A = M_iL_f$

Where $L_f$ is a constant known as heat of fusion and the value is $334*10^3 J/kg$

$Q_A = 0.129 *334 *10^3 = 43086 J$

The energy been released when the steam changes to water is mathematically given as

$Q_B = M_s * L_v$

Where $L_v$ is a constant known as heat of vaporization and the value is $2256*10^3J/kg$

$Q_B = 0.019 * 2256*10^3 = 42864J$

The energy released when the temperature of water decrease from 100°C to 0°C is

$Q_C = M_s *C_water ($ 100°C)

Where $C_{water}$ is the specific heat of water which has a value $4186J/kg \cdot K$

$Q_C = 0.019 *4186*100 = 7953.4$

Looking at the values we obtained we noticed that ]

$Q_B + Q_C > Q_A$

What this means is that the ice will melt

bearing in mind the conservation of energy

looking the way at which water at different temperature were mixed according to the question

Heat lossed by the vapor = heat gained by ice

$Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T$

$T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}$

$T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}$

$T = 12.48^oC$

3 0

3 years ago

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward

AleksandrR [38]

Answer:

(a) 104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.

The component of F perpendicular to the ramp is Fy.

(a)

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.

Resolve F into its x-component from Pythagorean theorem:

Fx=Fcos30°

Solve for F:

F= Fx/cos30°

Substitute for Fx from given data:

Fx=90 N/cos30°

=104 N

(b) Resolve r into its y-component from Pythagorean theorem:

Fy = Fsin 30°

Substitute for F from part (a):

Fy = (104 N) (sin 30°)

= 52 N

5 0

3 years ago

What phase difference between two identical traveling waves, moving in the same direction along a stretched string, results in t

Tom [10]

Answer:

Explanation:

Let the amplitude of individual wave be I and resultant amplitude be 1.703 I . Let the phase difference be Ф in terms of degree

From the formula of resultant vector

(1.703I)² = I² + I² + 2 I² cosФ

2.9 I² = 2I² + 2 I² cosФ

.9I² = 2 I² cosФ

cosФ = .9 / 2

= .45

Ф = 63.25 .

5 0

3 years ago