Momentum = 0.5 * 4 = 2
to conclude the man’s velocity after he throws the piece of equipment, divide
this number by the man’s mass.
v = 2/90
This is about 0.0222 m/s. To know if he can move 6 meters at velocity in
4minutes, use the following equation.
d = v * t, t = 4 * 60 = 240 s
d = 2/90 * 240 = 5⅓ meters.
This is ⅔ of a meter from the spaceship. To know the velocity that he must have
to move 6 meter, use the same equation.
6 = v * 240
v = 6/240
This is about 0.00416 m/s.
His final momentum = 90 * 6/240 = 2.25
To know the velocity of the package, divide this number by the mass of the
package.
v = 2.25/0.5 = 4.5 m/s
Answer:
Speed of the light in water= 225,000 km/s
Explanation:
At the speed with which light propagates through a homogeneous and transparent medium, it is a constant characteristic of that medium, and therefore, it changes from one medium to another.
Due to its enormous magnitude, the measurement of the speed of light has required the invention of ingenious procedures that will overcome the inconvenience of short land distances in relation to such extraordinary speed.
Astronomical methods and terrestrial methods have been giving ever closer results. At present, the value c = 299,792,458 km / s is accepted for the speed of light in a vacuum. In any transparent material medium the light propagates with a speed that is always lower than c. Thus, for example, in water it does so at around 75% of the speed of light in a vacuum: about 225,000 km / s.
Answer:
Pi(3.14) radians or 180º degrees
Explanation:
First of all, we need to obtain the wavelength of a wave traveling to the speed of sound and 420 Hz of frequency.
The formula is:

where l = wavelength in meters
With current values:
l = 336 [m/s]/420[1/s] = 0.8 meters
Since a complete cycle (360º or 2pi radians) needs 0.8 meters to complete, 0.4 meters or 40 cm is just half of it, making a 180º degree phase or 3.14 radians.
Answer:
i can see it and it looks really cool
Explanation:
Answer:
a) 1.6*10^6 V
b) 13.35*10^6 V
Explanation:
The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:
(1)
q1 = 3.90µC = 3.90*10^-6 C
q2 = -2.4µC = -2.4*10^-6 C
r1 = 1.25 cm = 0.0125 m
r2 = -1.80 cm = -0.018 m
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
You replace all the parameters in the equation (1):
![V=k[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\\\\V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0125m}+\frac{-2.4*10^{-6}C}{0.018m}]=1.6*10^6V](https://tex.z-dn.net/?f=V%3Dk%5B%5Cfrac%7Bq_1%7D%7Br_1%7D%2B%5Cfrac%7Bq_2%7D%7Br_2%7D%5D%5C%5C%5C%5CV%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0125m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.018m%7D%5D%3D1.6%2A10%5E6V)
hence, the total electric potential is approximately 1.6*10^6 V
b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:
r1 = 0.0150m - 0.0125m = 0.0025m
r2= 0.015m + 0.018m = 0.033m
Then, you replace in the equation (1):
![V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0025m}+\frac{-2.4*10^{-6}C}{0.033m}]=13.35*10^6V](https://tex.z-dn.net/?f=V%3D%288.98%2A10%5E9Nm%5E2%2FC%5E2%29%5B%5Cfrac%7B3.90%2A10%5E%7B-6%7DC%7D%7B0.0025m%7D%2B%5Cfrac%7B-2.4%2A10%5E%7B-6%7DC%7D%7B0.033m%7D%5D%3D13.35%2A10%5E6V)
hence, for y = 1.50cm you obtain V = 13.35*10^6 V