Answer:
A) Force = 0.69 N
B) Acceleration = 0.34 m/s^2
Explanation:
The electric force is given by the expression:
K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q1 and 12 is the charge of the particles, and r is the distance:
Part B.
For the acceleration, you need Newton's second Law:
F = m*a
Then,
One of two things is true about this question: EITHER it can't happen
as you've described it, OR you've left out some vital information.
-- IF the first stone was thrown downward with an initial speed and the
second one was dropped from rest 1 second later, then the second one
can never catch up with the first one, and they can never hit the water together.
-- IF the first stone was thrown downward with an initial speed, AND the
second one was released 1 second later, AND they actually do hit the
water together, THEN the second stone must have been given an initial
downward speed greater than 2 m/s, otherwise it could never catch up
with the first one.
Note:
The masses and weights of the stones are irrelevant and not needed.
=======================================================
An afterthought . . . . .
If the first stone was tossed UP at 2 m/s . . . that could be the meaning of the
prominent plus-sign that you wrote next to the 2 . . . then it rises for (2/9.8) second, then begins to fall, and passes the mountain climber's hand on the way down (4/9.8) second after he tossed it, falling at the same 2.0 m/s downward.
From there, it still has 50m to go before it hits the water.
50 = 2 T + 1/2 G T²
4.9 T² + 2 T - 50 = 0
T = 3 seconds
The first stone hits the water 3 seconds after passing the mountain climber's hand on the way down at a downward speed of 2.0 m/s. In that 3 seconds, it gains (3 x 9.8) = 29.4 m/s of additional speed, hitting the water at (29.4 + 2) = 31.4 m/s .
This is all just a guess, assuming that the 2.0 m/s was an UPWARD launch.
Maybe I'll come back later and calculate the second stone.
Explanation:
Substances become neutral at the number 7 because at this point, the acid and Base are equal and becomes neutralized
Answer:
the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr
Explanation:
Given:
speed of space vehicle =5000 km/hr
rocket motor speed = 71 km/hr relative to the command module
mass of module = m
mass of motor = 4m
By conservation of linear momentum
Pi = Pf
Pi= initial momentum
Pf= final momentum
Since, the motion is only in single direction
Where M is the mass of the space vehicle which equals the sum of motor's mass and the command's mass, Vi its initial velocity, V_mE is velocity of motor relative to Earth, and V_cE is its velocity of the command relative to Earth.
The velocity of motor relative to Earth equals the velocity of motor relative to command plus the velocity of command relative to Earth.
V_mE = V_mc+V_cE
Where V_mc is the velocity of motor relative to command this yields
substituting the values we get
= 4943.2 Km/hr
the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr
The restoring force in a simple harmonic oscillator acts as a restoring force, acting opposite in direction of and proportional in magnitude to the displacement. This has the result that when the oscillator reaches the equilibrium point, it has gained its maximum kinetic energy (and therefore velocity) while the magnitude of the restoring force, and therefore the acceleration, is 0.
Choice A