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Sever21 [200]
4 years ago
5

Suppose that a meter stick is balanced at its center.  A 0.24 kg mass is then positioned at the 6-cm mark.   At what cm mark mus

t a 0.35 kg mass be placed to balance the 0.24 kg mass?  (NEVER include units with the answer to ANY numerical question.)

Physics
1 answer:
Scilla [17]4 years ago
8 0

Answer:

80.17 cm

Explanation:

Taking moments of forces about the center, the total clockwise moments is equal to the total counter clockwise moment:

Force * distance (counter clockwise) = force * distance (clockwise)

0.24 * 9.8 * (50 - 6) = 0.35 * 9.8 * (x - 50)

0.24 * 44 = 3.43x - 171.5

103.5 = 3.43x - 171.5

=> 3.43x = 103.5 + 171.5

3.43x = 275

x = 275/3.43 = 80.17 cm

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Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistan
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Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a)  Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

Δy = 0.613 m

(b) Given in the x direction:

v₀ = 1.10 m/s

a = 0 m/s²

t = 0.350 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

vₓ = 1.10 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (10 m/s²) (0.350 s) + 0 m/s

vᵧ = 3.50 m/s

The magnitude is:

v² = vₓ² + vᵧ²

v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal

3 0
3 years ago
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