Answer:
The mass of the meter stick is 
Explanation:
Here since the meter stick is in equilibrium position its net torque and net force should be equal to zero
Since at the begging the meter stick is balanced at center of the meter stick that means its center of mass should be present at 
Now lets consider the later case where stick is balanced by two 5.12 g coins .
Here torque due to two coins = 
Torque due to weight of meter stick = 
= 
where m = mass of the meter stick
Here 
.
Upon equating we will be getting mass of the meter stick =
Answer:
3.5 N
Explanation:
Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm
- Tension of the string attached at the 0cm end is 0 as moment arm is 0
- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise
- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise
- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise
- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise
- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.
Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm
Total counter-clockwise moment = 100T
For this to balance, 100 T = 350
so T = 350 / 100 = 3.5 N
Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).
Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr. They'll only
change the units.
(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)
= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)
= 20 meter/second
Answer:
the number of grains in the ball is 274,848
Explanation:
Given that;
diameter = 0.5 mm
so radius r = 0.25 mm
first we determine the volume of the ball using the following equation;
V = 4/3×πr³
we substitute
V = 4/3×π(0.25)³
V = 0.06544 mm³
Now form table 1.1 "Grain sizes" a metal with grain size number of 12 has about 4,200,000 grains/mm³
so;
Number of grains N = 0.06544 × 4,200,000
N = 274,848 grains
Therefore, the number of grains in the ball is 274,848
