A small lab cart and one of larger mass collide and rebound off each other. Which of them has the greater average force on it du
1 answer:
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Answer:
q = 4.5 nC
Explanation:
given,
electric field of small charged object, E = 180000 N/C
distance between them, r = 1.5 cm = 0.015 m
using equation of electric field
k = 9 x 10⁹ N.m²/C²
q is the charge of the object
now,
q = 4.5 x 10⁻⁹ C
q = 4.5 nC
the charge on the object is equal to 4.5 nC
Answer:
82.76m
Explanation:
In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.
You use the formula for the circumference of the steel ring:
(1)
C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)
you solve for r in the equation (1):
Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:
(2)
L: final length of the tube = ?
Lo: initial length of the tube = 4*10^7m
ΔT = change in the temperature of the steel tube = 1°C
α: thermal coefficient expansion of steel = 13*10^-6 /°C
You replace the values of the parameters in the equation (2):
With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:
Finally, you compare both r and r' radius:
r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m
Hence, the distance to the ring from the ground is 82.76m
Answer:
2.5 m
Explanation:
Weight of billboard worker = 800 N
Number of ropes = 2
Length of scaffold = 4 m
Weight of scaffold = 500 N
Tension in rope = 550 N
The sum of the torques will be
The position of the person will be 2.5 m