<span>By pythagorean theorem then, the vertical side of the right triangle must be 12.
Then if x is the angle between the horizontal side and the hypotenuse, sin(x) = 12/13 but also the anser should be in this sentences.
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As per as my knowledge
The speed of a wave in a medium is affected by <u>d</u><u>e</u><u>n</u><u>s</u><u>i</u><u>t</u><u>y</u>,<u> </u><u>w</u><u>a</u><u>v</u><u>e</u><u>l</u><u>e</u><u>n</u><u>g</u><u>t</u><u>h</u> and <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u>:)
(Good luck on your test and mark me brainliest if this helps)
Answer:
diameter of largest orbit is 0.60 m
Explanation:
given data
isotopes accelerates KE = 6.5 MeV
magnetic field B = 1.2 T
to find out
diameter
solution
first we find velocity from kinetic energy equation
KE = 1/2 × m×v² ........1
6.5 × 1.6 ×
= 1/2 × 1.672 ×
×v²
v = 3.5 ×
m/s
so
radius will be
radius =
........2
radius =
radius = 0.30
so diameter = 2 × 0.30
so diameter of largest orbit is 0.60 m
W = mg, Assuming g ≈ 9.8 m/s² on the earth surface.
735 N = m* 9.8
735/9.8 = m
75 = m
Mass , m = 75 kg. B.
Answer:
145 m
Explanation:
Given:
Wavelength (λ) = 2.9 m
we know,
c = f × λ
where,
c = speed of light ; 3.0 x 10⁸ m/s
f = frequency
thus,

substituting the values in the equation we get,

f = 1.03 x 10⁸Hz
Now,
The time period (T) = 
or
T =
= 9.6 x 10⁻⁹ seconds
thus,
the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s
Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s
Now,
For radar to detect the object the pulse must hit the object and come back to the detector.
Hence, the shortest distance will be half the distance travelled by the pulse back and forth.
Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}
Thus, the minimum distance to target =
= 145 m