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3 years ago

If you wish to warm 100 kg of water by 20°C for your bath, how much heat is required? (Give your answer in calories and joules.)

1 answer:

4 0

Q = mcθ

Where m = mass of water in kg.
c = specific heat capacity in kJ/kg⁰C, c for water = 4200 kJ/kg⁰C
θ = temperature rise in ⁰C

Q = 100*4200* 20 Note here the temperature rise is 20 ⁰C
Q = 8 400 000 J

In calories, 4.2 J = 1 Calorie
= 8 400 000 / 4.2 = 200 000

Q = 200 000 Calories

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Why are Watson’s experiments considered controversial?

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D, it is considered unethical today

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3 years ago

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Based on the TIA/EIA 568-B structured cabling standard, the cabling that runs from the telecommunications closet to each work ar

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Answer:

False.

Explanation:

Backbone Cabling: A system of cabling that connects the equipment rooms and telecommunications rooms.

Horizontal Cabling: The system of cabling that connects telecommunications rooms to individual outlets or work areas on the floor.

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4 years ago

A rocket sled accelerates from 10 m/s to 40 m/s in 2 seconds. What is the average acceleration of the sled?

Dafna11 [192]

Answer:

15m/s²

Explanation:

Given parameters:

Initial velocity = 10m/s

Final velocity = 40m/s

Time taken = 2s

Unknown:

Average acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time;

Acceleration = $\frac{Final velocity - Initial velocity }{time}$

Acceleration = $\frac{40 - 10}{2}$ = 15m/s²

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3 years ago

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$

And the distance 'd' is

$d = lsin\theta$

$d = 1.14 sin 16.2\°$

$d = 0.318m$

Through the free-body diagram the tension components are given by

$Tcos\theta = mg$

$Tsin\theta = \frac{mv^2}{R}$

Here we can watch that,

$R = r+d$

Dividing both expression we have that,

$tan\theta = \frac{v^2}{Rg}$

Replacing the values,

$tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}$

$v = 4.83371m/s$

PART B) Using the vertical component we can find the tension,

$Tcos\theta = mg$

$T = \frac{mg}{cos\theta}$

$T = \frac{(10+26.2)(9.8)}{cos(16.2)}$

$T = 369.42N$

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4 years ago

A maple tree seed fell 180 centimeters straight toward the ground at a constant velocity. It moved that distance in 1.5seconds.

denis23 [38]

Answer: 1.2 m/s

Explanation:

Velocity $V$ is defined as the variation of position of an object or body in time. So, if we know the distance the seed traveled and the time, we can calculate its velocity: