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lesantik
11 days ago
10

2. On January 21 in 1918, Granville, North Dakota, had a surprising change in temperature. Within 12 hours, the temperature chan

ged from 237 K to 283 K. What is this change in temperature in the Celsius and Fahrenheit scales?
Physics
1 answer:
Yuliya22 [1.1K]11 days ago
7 0

Respuesta:

El cambio de temperatura en Celsius equivale a 46°C.

El cambio de temperatura en Fahrenheit equivale a 82.8°F.

Explicación:

Un grado Celsius es equivalente a un grado Kelvin; por lo tanto,

\Delta C = \Delta K = 283K-237K\\\\ \boxed{\Delta C = 46^oC}

Un grado Fahrenheit es 1.8 veces un grado Celsius; por lo tanto

\Delta F = 1.8(46^o)

\boxed{\Delta F = 82.8^oF}

Por lo tanto, el cambio en Celsius es de 46°C y en Fahrenheit es de 82.8°F.

You might be interested in
Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.
inna [987]
The heat required to raise the temperature of a substance by \Delta T is represented by
Q=m C_p \Delta T
where m stands for the mass of the substance and C_p indicates the specific heat of the substance. In this situation, we possess m=100~g=0.1~Kg and C_p=4.19~KJ/(Kg K), the specific heat of water.
Consequently, we can ascertain the temperature rise \Delta T:
\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C
Initially, the water's temperature was 25^{\circ}C, so the end temperature should be
T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ
where C_L denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of 100^{\circ}C (the temperature at which the phase change begins)

4 0
4 days ago
An object initially at rest experiences a constant horizontal acceleration due to the action of a resultant force applied for 10
inna [987]

Answer:

a = 18.28 ft/s²

Explanation:

the values provided are:

duration of force application, t= 10 s

Work done = 10 Btu

mass of the object = 15 lb

acceleration, a =? ft/s²

1 Btu = 778.15 ft.lbf

thus, 10 Btu = 7781.5 ft.lbf

m = \dfrac{15}{32.174}\ slug

m = 0.466 slug

So,

the work is equivalent to the change in kinetic energy

W = \dfrac{1}{2} m (v_f^2-v_i^2)

7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2

 v_f = 182.75\ ft/s

The acceleration of the object is therefore

  a = \dfrac{v_f-v_o}{t}

  a = \dfrac{182.75-0}{10}

         a = 18.28 ft/s²

the constant acceleration of the object is calculated to be 18.28 ft/s²

3 0
5 days ago
A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At
kicyunya [1025]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

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

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + \frac{1}{2}(2.25 x 15.4^{2})

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

v^{2} = u^{2} + 2gs

0^{2} = 34.65^{2} + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying v^{2} = u^{2} + 2gs

v^{2} = 0^{2} + 2(9.81 x 328)

v^{2} = 0 + 6435.36

v = \sqrt{6435.36} = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

7 0
5 days ago
A string is stretched by two equal but opposite forces f newton each what is tension in string
Maru [1056]

The string does not experience any force of tension, as it balances two forces acting in the same direction. Hence, the tension is zero.

Explanation:

If tension existed in the string, it would mean that two equal but opposite forces are exerting pull in contrary directions.

When a force of f newtons is applied from the right and another force of f newtons from the left, the resulting action occurs through one force. Because there is action on the same string in opposing directions, the tension in the string can only be equal to the magnitude of the string itself.

Therefore, the string indeed has no tension since it is dealing with two forces acting in the same direction. Thus, the tension is zero.

8 0
4 days ago
Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 3.2m long and
Softa [913]

Answer:

x = 0.29 m

Explanation:

It is known that the total external force acting on the mass system equals ZERO,

so the center of mass of the entire system will stay stationary.

We find that

m_1\Delta x_1 + m_2\Delta x_2 + m_3\Delta x_3 = 0

Since Ernie approaches Burt's position, we have:

33 x + 290 x + 32(-3.2 + x) = 0

355 x = 102.4

therefore, we conclude that

x = \frac{102.4}{355}

x = 0.29 m

8 0
9 days ago
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