how may kilojoules of heat are required to melt a 10.0g popsicle at 0 degrees C?(assume the same properties as water)

2 answers:

7 0

<span>there are different machines that help measure the distance of a property for construction workers to determine where to put a foundation </span>

Nikolay [14]3 years ago

3 0

The heat of fusion for water is 334j/g
therefore 1g of ice would require 334 x 10 j of energy which is 3.34kJ

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Help with these questions

erma4kov [3.2K]

<h2>Explanation:</h2><h3>3. </h3>

When light bounces back, it is <em>reflected</em>. (That's why you see your <em>reflection</em> in a mirror.) When light is bent from the path it is taking, it is <em>refracted</em>. The only answer choice that makes correct use of these terms is the third choice:

Part of the ray is <em>refracted</em> into ray B; part of the ray is <em>reflected</em> as ray R.

_____

The index of refraction is the ratio of the sine of the angle of incidence to the sine of the angle of refraction. Both angles are measured from the normal to the surface. The angle of refraction here is 12.5° less than the angle of incidence, 44°, so is 31.5°. Then the index of refraction of the medium is ...

n = sin(44°)/sin(31.5°) = 0.69466/0.52250 = 1.3299 ≈ 1.33

none of the offered choices is correct. The closest is 1.34.

8 0

3 years ago

Francium-223 has a half-life of 22 minutes. A Geiger counter gives a reading of 58 counts per minute (cpm) from a sample of fran

polet [3.4K]

Answer:

29 counts per minute (29 cpm)

Explanation:

The half-life of a radioactive isotope is the time needed for the activity of the sample to halve.The half-life of Francium-223 is exactly 22 minutes: this means that its activity after 22 minutes becomes exactly half of its initial value. Since the initial activity of this sample of francium-223 was 58 cpm, the activity after 22 minutes will be

$A=\frac{58 cpm}{2}=29 cpm$

8 0

3 years ago

. A rope is being used to pull a mass of 10 kg vertically upward. Determine the tension on the rope, if starting from rest, the

telo118 [61]

$\text{Given that,}\\\\\text{Mass, m =10 kg}\\\\\text{Time, t = 8 sec}\\\\\text{Velocity, v = 4~m/s}\\\\\text{When a body is moving upwards,}\\\\\text{Tension,}~ T=mg +ma\\\\~~~~~~~~~~~~~~~=mg+m\left(\dfrac{v-u}t \right)\\\\~~~~~~~~~~~~~~~=10(10)+10\left(\dfrac{4-0}8\right)\\\\~~~~~~~~~~~~~~~=100+10\left(\dfrac 12\right)\\\\~~~~~~~~~~~~~~~=100+5\\\\~~~~~~~~~~~~~~~=105~N$