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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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LuckyWell [14K]

Answer:

10.19 m

Explanation:

Water will rise to equalize the pressure inside and outside the tube.

The equation of pressure is given by

p=\rho gh

Where,

p = Pressure of air = 10⁵ N/m²

ρ = Density of water = 10³ kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Height the water will rise

h=\frac{p}{gh}\\\Rightarrow h=\frac{10^5}{9.81\times 10^3}\\\Rightarrow h=10.19\ m

∴ The water will rise by 10.19 m.

7 0
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A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
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barxatty [35]

Answer:

Velocity = 3.25[m/s]

Explanation:

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In point 1, (surface of the water) we have the atmospheric pressure and at point 2 the water is coming out also at atmospheric pressure, therefore this members in the Bernoulli equation could be cancelled.

The velocity in the point 1 is zero because we have this conditional statement "The water surface drops very slowly and its speed is approximately zero"

h2 is located at point 2 and it will be zero.

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