Answer:
0.54m
Explanation:
Step one:
given data
length of seesaw= 3m
mass of man m1= 85kg
weight = mg
W1= 85*10= 850N
mass of daughter m2= 35kg
W2= 35*10= 350N
distance from the center= (1.5-0.2)= 1.3m
Step two:
we know that the sum of clockwise moment equals the anticlockwise moment
let the distance the must sit to balance the system be x
taking moment about the center of the system
350*1.3=850*x
455=850x
divide both sides by 850
x=455/850
x=0.54
Hence the man must sit 0.54m from the right to balance the system
Take 68.2/60 = 1.137 hr
take 56.9/1.137 = 50.043 mi/hr
take 189/211 = 0.896
24.8/2 = 12.4 m
12.4/82.3 = 0.15s
The net force on particle particle q1 is 13.06 N towards the left.
<h3>
Force on q1 due to q2</h3>
F(12) = kq₁q₂/r₂
F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)
F(12) = -14.41 N (towards left)
<h3>Force
on q1 due to q3</h3>
F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)
F(13) = 1.352 N (towards right)
<h3>Net force on q1</h3>
F(net) = 1.352 N - 14.41 N
F(net) = -13.06 N
Thus, the net force on particle particle q1 is 13.06 N towards the left.
Learn more about force here: brainly.com/question/12970081
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Answer:
vf = 11.2 m/s
Explanation:
m = 10 Kg
F = 2*10² N
x = 4.00 m
μ = 0.44
vi = 0 m/s
vf = ?
We can apply Newton's 2nd Law
∑ Fx = m*a (→)
F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m
⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²
then , we use the equation
vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)
⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s