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Svetach [21]
2 years ago
7

Water flows through a cylindrical pipe of varying cross-section. The velocity is 3.0 m/s at a point where the pipe diameter is 1

.0 cm. What is the flow rate R
Physics
1 answer:
Setler79 [48]2 years ago
8 0

Answer:

The flow rate is  R =2.357 *10^{-4} \ m^3/s

Explanation:

From the question we are told that

    The velocity is v  =  3.0 \ m/s

   The  diameter of the pipe is  d =  1.0 \ cm  = 0.01 \ m

 

The  radius of the pipe is mathematically represented as

            r =  \frac{d}{2}

substituting values

            r =  \frac{0.01}{2}

           r =  0.005 \ m

The flow rate is mathematically represented as

       R  =  v  * A

Where is the cross-sectional area of the pipe which is mathematically evaluated as

      A   = \pi r^2

substituting values

      A   =  3.142 *  (0.005)^2

     A   = 7.855 *  10^{-5} \ m^2

So

    R  =  3.0  *  7.855 *10^{-5}

    R  =  2.357*10^{-4} \ m^3 /s

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Determine the kinetic energy of a 1000-kg roller coaster car that is moving with a speed of 20.0 m/s.
ankoles [38]

Answer:

The correct answer is - 200000 J

Explanation:

We use the formula of kinetic energy:

The formula to calculate kinetic energy is,

Here,

Ec =1/2 x  m x v^2

The mass of the roller coaster is,  m  =  1000  k g

The speed of the roller coaster is, v = 20.0 m/s

 Therefore,

Ec=1/2 x 1000kg x (20m/s)^2 = 200.000Joule

200,000J

7 0
3 years ago
as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p
miv72 [106K]

Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

E = (3)(230\ m/s)(0.61\ T)Sin90^o

<u>E = 420.9 N/C</u>

3 0
3 years ago
ate around its central axis. A rope wrapped around the drum of radius 1.24 m exerts a force of 4.56 N to the right on the cylind
Mashcka [7]

Answer:

Magnitude the net torque about its axis of rotation is 1.3338 Nm

Explanation:

The radius of the wrapped rope around the drum, r = 1.24 m

Force applied to the right side of the drum, F = 4.56 N

The radius of the rope wrapped around the core, r' = 0.57 m

Force on the cylinder in the downward direction, F' = 7.58 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

\tau = F\times r\tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'\tau_{net} = - 4.56\times 1.24 + 7.58\times 0.57 \\\\= - 1.3338\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 1.3338\ Nm

8 0
3 years ago
Answer all questions will mark as brainiest
Drupady [299]
11:C) sound travels fastest in solids
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7 0
3 years ago
A pendulum on earth oscillates with a period of 3.45 seconds. What is the length of the pendulum?
Sholpan [36]

Answer:

The length of the pendulum is 2.954 m.

Explanation:

Given;

period of the pendulum, T = 3.45 s

The period of the pendulum oscillation is given as;

T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{l}{g} }\\\\\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\l = \frac{gT^2}{4\pi ^2} \\\\

where;

L is length of the pendulum

g is acceleration due to gravity on Earth = 9.8 m/s²

l = \frac{gT^2}{4\pi ^2}\\\\l = \frac{(9.8)(3.45)^2}{4\pi ^2}\\\\l = 2.954 \ m

Therefore, the length of the pendulum is 2.954 m.

7 0
2 years ago
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