The number of significant digits to the answer of the following problem is four.
<h3>What are the significant digits?</h3>
The number of digits rounded to the approximate integer values are called the significant digits.
The following problem is
(2.49303 g) * (2.59 g) / (7.492 g) =
On solving we get
= 0.86184566204
The answer is approximated to 0.86185
Thus, the significant digits must be four.
Learn more about significant digits.
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Answer:

Explanation:
As per Doppler's effect we know that the frequency of the sound that is observed by the detector is the reflected sound
This reflected sound is given as


so we know that the beat frequency is

so we will have



so we have



Answer:
The knights collide 53.0 m from the starting point of sir George.
Explanation:
The equation for the position in a straight accelerated movement is as follows:
x = x0 + v0 t + 1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial speed
a = acceleration
t = time
The position of the two knights is the same when they collide. Since they start from rest, v0 = 0:
Sir George´s position:
xGeorge = 0 m + 0 m + 1/2 * 0.300 m/s² * t²
Considering the center of the reference system as Sir George´s initial position, the initial position of sir Alfred will be 88.0 m. The acceleration of sir Afred will be negative because he rides in opposite direction to sir George:
xAlfred = 88.0 m + 0 m - 1/2 * 0.200 m/s² * t²
When the knights collide:
xGeorge = x Alfred
1/2 * 0.300 m/s² * t² = 88.0 m - 1/2 * 0.200 m/s² * t²
0.150 m/s² * t² = 88.0 m - 0.100 m/s² * t²
0.150 m/s² * t² + 0.100 m/s² * t² = 88.0 m
0.250 m/s² * t² = 88.0 m
t² = 88.0 m / 0.250 m/s²
t = 18.8 s
At t = 18.8 s the position of sir George will be
x = 1/2 * 0.300 m/s² * (18.8 s)² = <u>53.0 m </u>
Answer:
19.6 cm.
Explanation:
From the question given above, the following data were obtained:
Focal length (f) = 13.8 cm
Magnification (M) = +2.37
Object distance (u) =.?
Next, we shall determine the image distance. This can be obtained as follow:
Magnification (M) = +2.37
Object distance (u) = u
Image distance (v) =?
M = v / u
2.37 = v / u
Cross multiply
v = 2.37 × u
v = 2.37u
Finally, we shall determine the object distance. This can be obtained as follow:
Focal length (f) = 13.8 cm
Image distance (v) = 2.37u
Object distance (u) =.?
1/v + 1/u = 1/f
vu / v + u = f
2.37u × u / 2.37u + u = 13.8
2.37u² / 3.37u = 13.8
Cross multiply
2.37u² = 3.37u × 13.8
2.37u² = 46.506u
Divide both side by u
2.37u² / u = 46.506u / u
2.37u = 46.506
Divide both side by 2.37
u = 46.506 / 2.37
u = 19.6 cm
Thus, the lens should be held at a distance of 19.6 cm.