Answer:
<h3>1.03684m</h3>
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g where
R is the distance moves in horizontal direction = 18.4m
H is the height
U is the velocity of the baseball = 40m/s
g is the acceleration due to gravity = 9.8m/s²
Substitute the given parameters into the formula and calculate H as shown;
18.4 = 40√2H/9.8
18.4/40 = √2H/9.8
0.46 = √2H/9.8
square both sides;
(0.46)² = (√2H/9.8)²
0.2116 = 2H/9.8
2H = 9.8*0.2116
2H = 2.07368
H = 2.07368/2
H = 1.03684m
Hence the ball is 1.03684m below the launch height when it reached home plate.
Answer:
In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.
Insulin and glucagon help maintain blood glucose levels
Answer:
Explanation:
E=(σ/ε0)
As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."
Answer:
18,850 Hz
Explanation:
We need to figure out the wavelength of the sound wave.
Thus,
Wavelength = 1000 * Lowest Amplitude Wave
Wavelength = 1000 * 2.0 * 10^(-5)
Wavelength = 0.02
Or,
Now, we need the frequency of this wave. It goes by the formula:
Where
f is the frequency in Hz
v is the speed of sound in air (to be 377 m/s)
is the wavelength (we found to be 0.02)
Substituting, we find the frequency:
The wave has frequency of 18,850 Hz
