The work done is calculated by multiplying the Force by the distance. So, 10x20 = 200 joules.
I hope this is the answer you are looking for, if not good luck on finding your answer ;D
Answer:
M₂ = M then L₂ = L
M₂> M then L₂ = \frac{M}{M_{2}} L
Explanation:
This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive
∑ τ = 0
The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂
The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.
M L + M₁ 0 - m₂ L₂ = 0
M L - m₂ L₂ = 0
L₂ =
L
From this answer we have several possibilities
* if the two masses are equal then L₂ = L
* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L
Please solve it ,,,,,,,,,,................................
fredd [130]
Answer:
the distance covered by the body in 5th second =u+a(n- 1/2)=7+4(5- 1/2)
=7+4×9/2
=7+18
=25m
Answer:
0.557 s
Explanation:
Given:
v₀ = 5.46 m/s
v = 0 m/s
a = -9.8 m/s²
Find: t
v = at + v₀
0 m/s = (-9.8 m/s²) t + 5.46 m/s
t = 0.557 s
Answer:
a) 4500 cycles b) 0.0667s c) 6.67s
Explanation:
a) 15 Hz= 15 cycles/ s
5 mins= 300s
15 cycles/s * 300s= 4500 cycles
b) Period= 1/ frequency
Period= 1/ 15 cycles/s
Period= 0.0667s
c) Period * number of revolutions= time
0.0667 * 100= 6.67s