Radiation is a type of heat transfer wherein there is no need for medium or media through which the heat will flow. Consequently, the radiation waves are able to travel through vacuum. The best observation as evidence to conclude that heat is indeed transferred by radiation is the increase of temperature of the receiving body.
Answer:
the answer is b
Explanation:
Second and third class levers are differentiated by <u>the location of the </u><u>load.</u>
<em>Hope</em><em> </em><em>this</em><em> </em><em>help</em><em> </em><em>you</em><em> </em><em>out </em><em>and have</em><em> </em><em>a </em><em>nice</em><em> </em><em>day </em><em>=</em><em>)</em>
If you do not have to use relative physics but classic physics, this is how you solve it:
Speed of light = c = 3 * 10^5 km/s
Speed of your foe respect to you: 0.259c
Speed of the torpedo respect to you: 0.349c
Speed of the torpedo respect your foe: 0.349c - 0.259c = 0.09c
Conversion to km/s = 0.09 * 3.0 * 10^5 km/s = 27000 km/s
Note that this solution, using classic physics do not take into account time and space dilation.
Answer: 27000 km/s
The orbiting velocity of the satellite is 4.2km/s.
To find the answer, we need to know about the orbital velocity of a satellite.
<h3>What's the expression of orbital velocity of a satellite?</h3>
- Mathematically, orbital velocity= √(GM/r)
- r = radius of the orbital, M = mass of earth
<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>
- M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
- Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)
=4.2km/s
Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.
Learn more about the orbital velocity here:
brainly.com/question/22247460
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Answer:
Option B
Explanation:
Magnification of Microscope is
Mo= Magnification of objective lens and
Me= magnification of the eyepiece.
Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.
Magnification,
when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.
Thus. Magnification will increase by decreasing the focal length.
The correct answer is Option B i.e. using shorter focal length
