Answer:
100 cc
Explanation:
Heat released in cooling human body by t degree
= mass of the body x specific heat of the body x t
Substituting the data given
Heat released by the body
= 70 x 3480 x 1
= 243600 J
Mass of water to be evaporated
= 243600 / latent heat of vaporization of water
= 243600 / 2420000
= .1 kg
= 100 g
volume of water
= mass / density
= 100 / 1
100 cc
1 / 10 litres.
No.
Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.
<h3>What is a random error?</h3>
Random error is defined as the deviation of the total error from its mean value due to chance.
Random errors can result from the instrument not being precise or from mistakes by the researcher.
Random errors can be minimized by taking multiple readings and averaging the results.
Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a ransom error has been minimized.
Learn more about random errors at: brainly.com/question/22041172
<span>D) Electromagnetic radiation travels in the form of longitudinal waves.</span>
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Below are the choices:
only the x component
only the y component
both the x and y components
neither the x nor the y component
The answer is neither the x nor the y component
Answer:
E
= -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E
= -4556.18 N/m
