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andrew11 [14]
2 years ago
14

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

ed of 16 m/s at an angle 32 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
1) What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
2) What is the vertical component of the ball’s velocity when it leaves Julie's hand?
3) What is the maximum height the ball goes above the ground?
4) What is the distance between the two girls?
5) How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)
Physics
1 answer:
iris [78.8K]2 years ago
3 0

Answer:

Explanation:

1.  V_{x} = V_{0} * cos\alpha ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. V_{y} = V_{0} * sin\alpha ⇒ 16* sin32 ≈ 9.4 m/s

3. y_{max} = \frac{v_{0}^2*sin^2\alpha}{2g}= \frac{16^2*sin^232}{2*9.8} (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

y_{max} ≈ 3.6677+1.5 ≈ 5.2m

4.  x_{max} = \frac{v_{0}^2*sin(2\alpha)}{g}=\frac{16^2*sin(2*32)}{9.8} ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5

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Answer:

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a)

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d = 50.2 m + 50.2 m = 100.4 m

The time can be found dividing the distance by the velocity:

t1 = 50.2 m / 2.21 m/s = 22.7149 s

t2 = 50.2 m / 4.11 m/s = 12.2141 s

t = t1 +t2 = 34.9290 s

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b)

Here we can easily know the total time:

t = 1 min + 1.16 min = 129.6 s

Now the distance wil be found multiplying each velocity by the time it has travelled:

d1 = 2.21 m/s * 60 s = 132.6 m

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Answer:

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E(r) = \frac{kp}{r^{3} } , where p = q × s

E(r) = \frac{kqs}{r^{3} } where r>>s

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F=QE

F= Q(\frac{kqs}{r^3})

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

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τ = F × r

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