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Sati [7]
3 years ago
13

If a ball rolls off a cliff that is 40.0 meters tall and lands 10.0 meters from the base of the cliff, with what initial horizon

tal velocity did the ball roll off the cliff? Neglect air resistance.
A.2.80 m/s
B.3.00 m/s
C.3.50 m/s
D.4.20 m/s
Physics
1 answer:
IrinaVladis [17]3 years ago
6 0

Option (c) is correct.

Explanation:

initial velocity along vertical=0

velocity along horizontal Vx

acceleration along the vertical=9.8 m/s²

displacement along vertical,h=40 m

distance along horizontal= 10 m

using the kinematic equation

h= vi*t + 1/2 g t²

40=0(t) + 1/2 (9.8)t²

t²=8.163

t=2.857 s

now the displacement along vertical= x= Vx* t

10=Vx (2.857)

Vx=3.5 m/s

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The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2 
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w1 = 17.94 or 18 rad s^-1 
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X-rays
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6 0
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List two methods by which a community can save water. In your own words, explain how these methods would help to increase the ye
Oliga [24]
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3 years ago
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labe
spin [16.1K]

Answer:

1.19 m/s²

Explanation:

The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so

f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀    (1)

Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²

Equating (1) and (2) we ave

2(√mg/μ)/f = T²g/4π²

Making g subject of the formula

g = 2π√(2√(m/μ)/f)/T

The period T = 316 s/100 = 3.16 s

Substituting the other values into , we have

g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16

g = 2π√(2 × 35.877/200 Hz)/3.16

g = 2π√(71.753/200 Hz)/3.16

g = 2π√(0.358)/3.16

g = 2π × 0.599/3.16

g = 1.19 m/s²

6 0
3 years ago
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lina2011 [118]

Answer:

500N/M

Explanation:

given that

force=10N

distance=50M

moment=force*distance

=10×50=500j

8 0
2 years ago
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