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Korvikt [17]
3 years ago
9

A ball is thrown horizontally from the top of a building at 2 m/s. It takes 3 seconds to reach the ground. How far did the ball

travel horizontally (dx)?
Physics
1 answer:
svlad2 [7]3 years ago
4 0
Every second it travels 2 meters and it traveled 3 so (2 x 3) would be 6meters it traveled
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How much time does it take a power drill accelerating at 64.3 rad/s2 to achieve
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Answer;

= 0.244 seconds

Explanation;

500 rpm is equivalent to; 1500 × 2π radians per minute

   = 9424.8 rad/minute  

To get revolutions per second we divide by 60

  = 9424.8/60  

   = 157.08 radians per second.  

Then we divide by 64.3 rad/s^2 to get time;

    = 157.08/64.3

     = 0.244 seconds.

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3 years ago
A horizontal disk with a radius of 23 m rotates about a vertical axis through its center. The disk starts from rest and has a co
lys-0071 [83]

Answer:

time is 0.42 sec

Explanation:

Given data

radius = 23 m

angular acceleration = 5.7 rad/s²

to find out

time

solution

we know that radius is constant so that

tangential acceleration At = angular acceleration × radius   ............. 1

tangential acceleration =  5.7 × 23 = 131.1 m/s²

and

radial acceleration Ar =  (angular velocity)² × radius    ........................2

we consider angular velocity = ω

this is acting toward center

so

compare 1 and 2

At = Ar

5.7 r =ω³ r

ω = √5.7 = 2.38746 rad/s

so

ω = 5.7 t

2.387 = 5.7 t

t =  2.387 / 5.7

t = 0.4187

time is 0.42 sec

8 0
3 years ago
Oil having a specific gravity of 0.9 is pumped as illustrated with a water jet pump. The water volume flowrate is 1 m3 /s. The w
love history [14]

Answer:

The rate at which the pump moves oil is 1 m³/s

Explanation:

Assumptions:

  • there is steady-state flow
  • oil and water are incompressible
  • first fluid is water, second fluid is oil and third fluid is the mixture of oil and water.

\rho_1Q_1 + \rho_2Q_2 = \rho_3Q_3 -------equation (i)

where;

ρ is the fluid density

Q is the volumetric flow rate

Q_1 + Q_2 = Q_3--------equation (ii)

Substitute in Q₃ in equation i

\rho_1Q_1 + \rho_2Q_2 = \rho_3(Q_1 +Q_2)

divide through by ρ₁

\frac{\rho_1Q_1}{\rho_1}+ \frac{\rho_2Q_2}{\rho_1} =\frac{ \rho_3(Q_1 +Q_2)}{\rho_1}\\\\Note; \frac{\rho_2}{\rho_1} = \gamma_2 \ and \ \frac{\rho_3}{\rho_1} = \gamma_3\\\\Q_1 + \gamma_2Q_2 = \gamma_3(Q_1+Q_2)

Make Q₂ the subject of the formula

Q_2 = \frac{Q_1(1- \gamma_3)}{\gamma_3-\gamma_2} = \frac{1 (\frac{m^3}{s}) (1-0.95)}{0.95-0.9} = 1 \ \frac{m^3}{s}

Therefore, the rate at which the pump moves oil is 1 m³/s

6 0
3 years ago
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