The ball's gravitational potential energy is converted into kinetic energy as it falls toward the ground.
<h3>How can the height of a dropped ball be determined?</h3>
Y = 1/2 g t 2, where y is the height above the ground, g = 9.8 m/s2, and t = 1.3 s, is the formula for problems like these. Any freely falling body with an initial velocity of zero meters per second can use this formula. figuring out how much y is.
A ball drops from the top of a building and picks up speed as it descends. Its speed is increasing by 10 m/s every second. What we refer to as motion with constant acceleration is, for example, a ball falling due to gravity.
The ball's parabolic motion causes it to move at a speed of 26.3 m/s right before it strikes the ground, which is faster than its straight downhill motion, which has a speed of 17.1 m/s. Take note of the rising positive y direction in the above graphic.
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From Boyle's law, the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature.
Therefore; P1V1 =P2V2; where PV is a constant
hence; 12 × 6 = 3× p2
p2 = 72/3
= 24 atm
Therefore; the new pressure will be 24 atm
It could rotate while not advancing distance
Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T
Answer:
B 14.5 m/s to the east
Explanation:
We can solve this problem by using the law of conservation of momentum.
In fact, if the system is isolated, the total momentum of the system must be conserved.
Here the total momentum before the stuntman reaches the skateboard is:
where
M = 72.0 kg is the mass of the stuntman
v = 15.0 m/s is his initial velocity (to the east)
The total momentum after the stuntmen reaches the skateboard is:
where
m = 2.50 kg is the mass of the skateboard
v' is the final velocity of the stuntman and the skateboard
Since momentum must be conserved, we have
And solvign for v',
And since the sign is the same as v, the direction is the same (to the east).
