Convert the number from scientific into standard notation: 5.9 x 10-2

1 answer:

6 0

Move the decimal point to:
Left : (if the exponent of ten is a negative number -) ... OUR CASE HERE (-2)
or to
Right : (if the exponent is positive +).

You should move the point as many times as the exponent indicates.
Do not write the power of ten anymore.

So, standard form is:
Two points to the left {Exponent of Ten is Negative (-2)}
0.059 ... (without the 10)

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A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear

Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

$f_o = f_s\frac {(v+vs)}{v}$ and $f_o = f_s\frac {(v-vs)}{v}$ when moving towards observer

With $f_o$ of 76 then taking speed in air as 343 m/s we have

$76 = f_s\times\frac {(343-vs)}{343}$

$f_s=\frac {343\times 76}{343-v_s}$

Similarly, with $f_o$ of 65 we have

$65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}$

Now

$f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}$

v_s=27.76 m/s

Substituting the above into any of the first two equations then we obtain

$f_s=70.07 Hz$

4 0

2 years ago

30. How do you make a conclusion?

FromTheMoon [43]

B. You compare the data

6 0

2 years ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

pickupchik [31]

The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.

In that particular situation, you can prove it like this: 

initial velocity is Vo 
launch angle is α 

initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity is 
Vh = Vo×cos(α) 

total time in the air is the the time it needs to fall back to a height of 0 m, so 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time = ? 
a = acceleration by gravity = g (= -9.8 m/s²) 
so 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now look at the horizontal range. 
r = v × t 
where 
r = horizontal range = ? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
so 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°

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2 years ago

Iodine 131 half life is 8.0 days. Ten percent of the original sample o his isotope remains after (a) 22.7 days (b) 24.9 days (c)

Artyom0805 [142]

Answer:

option (c) is correct

Explanation:

Half life of a substance is the time in which the element becomes half of is initial value.

half life, T = 8 days

Amount remaining, N = 10 % of original value

Let the original value is No.

N = 10% of No