All categories

4 years ago

How to calculate thermal energy

1 answer:

6 0

Heat, a measure of thermal energy, can be transferred from one point to another. Heat flows from the point of higher temperature to one of lower temperature. The heat content, Q, of an object depends upon its specific heat, c, and its mass, m. The Heat Transfer is the measurement of the thermal energy transferred when an object having a defined specific heat and mass undergoes a defined temperature change.

\Hope this helps/

You might be interested in

Write in point-slope form an equation of the line that passes through the point (1, 4) with slope -2.

Elan Coil [88]

Answer:

y-4=-2(x-1)

Step-by-step explanation:

hope this helped :)

3 0

3 years ago

The length of a rectangle is 5 centimeters less than twice its width. The perimeter of the rectangle is 80 cm. What are the dime

Sonbull [250]

Hello,

Answer A
Let's the width of the rectangle

2a-5 its length
Teh half perimeter is 80/2=40
a+2a-5=40 ==>3a=45==>a=15 and 2a-5=30-5=25

5 0

3 years ago

Read 2 more answers

Order these numbers from least to greatest 1/2 1/5 1/3

Ulleksa [173]

Answer:

least to greatest:

1/3/21/51

Step-by-step explanation:

4 0

3 years ago

A bird has a mass of 26 g and perches in the middle of a stretched telephone line. (a) Show that the tension in the line can be

Semmy [17]

Answer:

a) is demonstrated below, b) T=1.462N, c) T=14.6N

Step-by-step explanation:

a) Refer to the attached diagram.

Since the bird is standing in the middle of the line and each half is a straight line, Ta=Tb, so we will call the tension T=Ta=Tb and Tay=Tby

By trigonometry Tay=Ta·Sinθ

Since the system is in equilibrium W=Tay+Tby then:

W=2·Tay=2·Ta·Sinθ=2·T·Sinθ

Since W=mg, being m the mass of the bird and g, gravity:

$mg=2TSin\theta$

Isolating T, we demonstrate that

$T=\frac{mg}{2Sin\theta}$

b) Replacing θ=5º, m=0.026kg and g=9.8m/s² in the last equation, we can get the tension in Newtons:

$T=\frac{0.026*9.8}{2Sin5}=1.462N$

c) With θ=0.5º

$T=\frac{0.026*9.8}{2Sin0.5}=14.6N$

7 0

3 years ago

A poll found that 58% of U.S. adult Twitter users get at least some news on Twitter. The standard error for this estimate was 2.

Radda [10]

Answer: (53.49%, 62.51%)

Step-by-step explanation:

Given : A poll found that 58% of U.S. adult Twitter users get at least some news on Twitter.

i.e. p= 0.58

The standard error for this estimate was 2.3%

i.e. S.E.=0.023

Critical value for 95% confidence : $z_{\alpha/2}=1.96$

Confidence interval : $p\pm z_{\alpha/2}(S.E.)$

i.e. $0.58\pm (1.96)(0.023)$

$=0.58\pm 0.04508=(0.58-0.04508,\ 0.58+0.04508)\\\\=(0.53492,\ 0.62508)=(53.492\%,\ 62.508\%)=(53.49\%,\ 62.51\%)$

Conclusion : We are 95% confident that the true population proportion of U.S. adult Twitter users who get some news on Twitter falls in interval (53.49%, 62.51%)

5 0

4 years ago