<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>
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A 10.0 cm3 sample of copper has a mass of 89.6
1 answer:
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A. Formula: F=ma or F/m=a
10,000N/1,267kg≈7.9m/
B. Formula: a= and s=d/t
speed= 394.6/15
s=26.3m/s
a=
a=1.75m/
C. 7.9-1.75=difference of 6.15m/
D. The force that most likely caused this difference is friction forces
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