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rosijanka [135]
2 years ago
11

150J of heat energy

Physics
2 answers:
arsen [322]2 years ago
6 0

Btu/(lb-°F) J/(g-°C i mean this is the correct answer

4vir4ik [10]2 years ago
4 0

Answer:

<u>0.300 J/(g*C)</u>

Explanation:

Specific heat has a unit that describes the amount of heat (J) is required to raise a unit mass (g) by 1 degree temperature (C),  It should take the form of J/(g*C) if we stick with the metric units provided in the question.  Be aware that specific heat can be written in many different formats, all equivalent, but with differing measures.  Examples:

(BTU/kgC)

kJ/(mole*K)

J/(g*C)

and so on.  The specific heat for a material can be expressed in any of these formats.  There are all equivalent, just expressed with different units.

--------------

In this problem:  (150J)/[(100g)*(5C)] = 0.300 J/(g*C)

An accurate value of specific heat may be used for identification of the metal.  The closest I found is Barium (0.290 J/(g*C) and Zinc (0.388 J/(g*C)

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Reptile [31]

Answer:

Its inductance L = 166 mH

Explanation:

Since a current, I = 0.698 A is obtained when a voltage , V = 5.62 V is applied, the resistance of the coil is gotten from V = IR

R = V/I = 5.62/0.698 = 8.052 Ω

Since we have a current of I' = 0.36 A (rms) when a voltage of V' = 35.1 V (rms) is applied, the impedance Z of the coil is gotten from

V₀' = I₀'Z where V₀ = maximum voltage = √2V' and I₀ = maximum current = √2I'

Z = V'/I' = √2 × 35.1 V/√2 × 0.36 V = 97.5 Ω

WE now find the reactance X of the coil from

Z² = X² + R²

X = √(Z² - R²)

= √(97.5² - 8.05²)

= √(9506.25 - 64.8025)

= √9441.4475

= 97.17 Ω

Now, the reactance X = 2πfL where f = frequency of generator = 93.1 Hz and L = inductance of coil.

L = X/2πf

= 97.17/2π(93.1 Hz)

= 97.17 Ω/584.965 rad/s

= 0.166 H

= 166 mH

Its inductance L = 166 mH

5 0
3 years ago
21. Calculate the acceleration of the bus from point D to E. Show your work.
Marat540 [252]

21) Acceleration from D to E: 1 m/s^2

22) The acceleration of the bus from D to E is 1 m/s^2

Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time elapsed

In this problem, we want to measure the acceleration of the bus from point D to point E. We have:

- Initial velocity at point D: u = 0

- Final velocity at point E: v = 5 m/s

- Time elapsed from D to E: t = 21 - 16 = 5 s

Therefore, the acceleration between D and E is

a=\frac{5-0}{5}=1 m/s^2

22) This question is the same as 21), so the result is the same.

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4 0
3 years ago
If the area of an iron rod is 10 cm by 0.5 cm and length is 35 cm. Find the value of resistance, if 11x10^-8 ohm.m be the resist
malfutka [58]

Answer:

Resistance of the iron rod, R = 0.000077 ohms    

Explanation:

It is given that,

Area of iron rod, A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2

Length of the rod, L = 35 cm = 0.35 m

Resistivity of Iron, \rho=11\times 10^{-8}\ \Omega-m

We need to find the resistance of the iron rod. It is given by :

R=\rho\dfrac{L}{A}

R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}

R=0.000077 \Omega

So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.

6 0
3 years ago
Part of the plant that isresponsible for the transport of water from the roots to the leaves​
nydimaria [60]

Answer:

if you mean *responsible* for the transport of water from the roots to leaves is Xylem

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What is the minimum runway length that will serve?

Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.

A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.

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6 0
2 years ago
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