A 1000-kg space probe is motionless in space. to start moving, its main engine is fired for 5 s during which time it ejects exha
1 answer:
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Jogger moves in three displacements
d1 = 10 blocks East
d2 = 5 blocks South
d3 = 2 blocks East
now we can say
total displacement towards East direction will be
Total displacement towards South
now to find the net displacement we can use vector addition
<em>so magnitude of net displacement will be equal to 13 blocks</em>
Answer:
See the explanation below.
Explanation:
12) When an object is falling, how does the objects velocity change? what formula do you use?
The speed of a falling object is increased by a value of 9.81 meters per second per second. That is if we throw any body regardless of mass from a considerable height, its speed in the first second will be 9.81[ m/ s] , in the next second will be equal to 19.62 [m/s] in the next will be equal to 29.43 [m/ s].
The formula is:
where:
vo = initial velocity = 0
g = gravity = 9.81[m/s^2]
t = time [s]
13)
what is a falling speed at 6s, 9s, 112s?
v = 0 + (9.81*6) = 58.86[m/s]
v = 0 + (9.81*9) = 88.29 [m/s]
v = 0 + (9*112) = 1098.72 [m/s]
14)
If an object is falling at 65 [m/s]. How long has it been falling ? what is the formula that you use?
The formula is the same:
65 = 0 + 9.81*t
t = 65/9.81
t = 6.62[s]
15)
What formula is used to determine the distance an object is falling ?
where:
y = distance [m]
yo = initial distance, in most of the cases and depending of the reference point it will be eqaul to zero
vo = initial velocity, if it is free fall, then = 0
t = time [s]
g = gravity = 9.81[m/s^2]
This equation will be reduce to:
16)
using the times given in problem 13. Determine the distance fallen for each.
y = 0.5*9,81*(6)^2 = 176.58 [m]
y = 0.5*9,81*(9)^2 = 397.3 [m]
y = 0.5*9,81*(112)^2 = 61528.3 [m]
17)
If an object has fallen a distance of 87.3 [m]. How long was it falling?
87.3 = 0.5*9.81*t^2
Answer:
Explanation:
The electric field inside a parallel plate capacitor is
where A is the area of one of the plates, and Q is the charge on the capacitor.
The electric force on the electron is
where q is the charge of the electron.
By definition the capacitance of the capacitor is given by
Plugging this identity into the force equation above gives
The work done by this force is equal to change in kinetic energy.
W = Fx = (30q)(0.05) = 1.5q = K
The charge of the electron is
Therefore, the kinetic energy is