Answer:34 cm
Explanation:
Given
mass of meter stick m=80 gm
stick is balanced when support is placed at 51 cm mark
Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark
balancing torque
Answer:
15,505 N
Explanation:
Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student
-ΔU = ΔK
-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²
-(0 - mgh) = 1/2mv² - 0
mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.
mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 1 m)
v = √(19.6 m²/s²)
v = 4.43 m/s
Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s
So Ft = m(v' - v)
F = m(v' - v)/t
substituting the values of the variables, we have
F = 70 kg(0 m/s - 4.43 m/s)/0.02 s
= 70 kg(- 4.43 m/s)/0.02 s
= -310.1 kgm/s ÷ 0.02 s
= -15,505 N
So, the force transmitted to her bones is 15,505 N
Answer:
<em>a) Fvt cosθ</em>
<em>b) Fv cosθ</em>
<em></em>
Explanation:
Each horse exerts a force = F
the rope is inclined at an angle = θ
speed of each horse = v
a) In time t, the distance traveled d = speed x time
i.e d = v x t = vt
also, the resultant force = F cosθ
Work done W = force x distance
W = F cosθ x vt = <em>Fvt cosθ</em>
<em></em>
b) Power provided by the horse P = force x speed
P = F cosθ x v
P = <em>Fv cosθ</em>
Answer:
1.05 N
Explanation:
K = 0.7 N/m
e = 1.5 m
F = ?
from Hooke's law of elasticity
F = Ke
= 0.7×1.5
= 1.05 N
