A 70kg man spreads his legs as shown calculate the tripping force​

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A 40 kg child throws a 0.5 kg stone at a velocity of 5 m/s. To find the recoil, we apply the conservation of momentum formula: m1•v1 + m2•v2 = 0, where m1 is the mass of the child, and v1 is the child's recoil velocity. Applying the known values results in 40•v1 = -0.5 × 5, leading to v1 = -2.5 / 40, which simplifies to v1 = -0.0625 m/s. Thus, the child's recoil speed is 0.0625 m/s.

Answer:

(A) = 3.57 m

Explanation:

According to the question, the information provided is:

diameter (d) = 3.2 m

mass (m) = 42 kg

angular speed (ω) = 4.27 rad/s

Using the conservation of energy principle, we have

mgh = 0.5 mv² + 0.5Iω²...equation 1

where

Inertia (I) = 0.5mr²

ω = v/r

Revising equation 1, it turns into

mgh = 0.5 mv² + 0.5(0.5mr²)(v/r)²

resulting in gh = 0.5 v² + 0.5(0.5)v²

This simplifies to 4gh = 2v² + v²

thus h = 3v² ÷ 4g... equation 2

Given ω = v/r, we find v = ωr = 4.27 × (3.2 ÷ 2)

which yields v = 6.8 m/s

Next, substituting the value of v into equation 2 gives us

h = 3v² ÷ 4g

h = 3 × (6.8)² ÷ (4 × 9.8)

h = 3.57 m

Answer:

2023857702.507 m

Explanation:

Using Newton's law of gravitation:

G = gravitational constant

m_shrew = 50 g

m_elephant = 5 × 10^3 kg

r_earth = Earth's radius, 6400 km or 6,400,000 m

m_earth = Earth's mass

Equate the gravitational forces:

G m_shrew m_earth / r_earth^2 = G m_elephant m_earth / r^2

Cancel common terms on both sides:

m_shrew / r_earth^2 = m_elephant / r^2

Rearranged to solve for r^2:

r^2 = (m_elephant × r_earth^2) / m_shrew

Substituting the values:

r^2 = 4.096 × 10^{13}

Taking square root gives:

r = 2,023,857,702.507 m

Answer:

h = 1/50 v₀² g

Explanation:

This scenario illustrates conservation of momentum; the system comprises two bodies, hence forces during the collision are internal and momentum remains conserved.

Prior to the collision

   po = 0 + (m/4) vo

Post collision

     pf = (m + m/4) v

     pf = (5m/4) v

     m/4 vo = (5m/4) v

     v = 1/5 vo

This indicates that after impact, the bodies travel at 1/5 of their initial velocity.

To determine the maximum height, we will apply the law of energy conservation.

At the peak post-collision

     Em = K = ½ (5m / 4) v²

At the apex

    Em = U = (5m/4) g h

   ½ (5m/4) v² = (5m/4) g h

    h = ½ v² / g

    h = ½ (1/5 v₀)² / g

    h = 1/50 v₀² g