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sergejj [24]
2 years ago
15

Define reference point ​

Physics
1 answer:
Fynjy0 [20]2 years ago
8 0

Answer:

A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point.

Hope this helps:)

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What real world examples show no work being done? Can you think of examples other than resisting the forces of gravity?
Lesechka [4]
I need more info man then i can help you. 
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3 years ago
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A disk of radius 25 cm spinning at a rate of 30 rpm slows to a stop over 3 seconds. what is the angular acceleration?
SCORPION-xisa [38]

Answer:

α = -π/3 rad/s²

θ = 1.5π rad ≈ 4.71 rad

θ = 0.75 rev

Explanation:

30 rev/min (2π rad/rev) / (60 s/min) = π rad/s

α = (ωf - ωi) / t = (0 - π) / 3 = -π/3 rad/s²

θ = ½αt² = ½(π/3)3² = 1.5π rad ≈ 4.71 rad

θ = 1.5π rad / 2π rad/rev = 0.75 rev

5 0
3 years ago
Heres what <br> GIMBAP KIMCHI PORK BELLY look like
Molodets [167]

Answer:

OWOWOWOOWOWOWEO. PORK = PORK NOW GIOVE BRAINLIEST BC I SAID PORK = PORK

Explanation:

3 0
3 years ago
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Water at 20 C flows through a 5-cm-diameter pipe that has a 180 vertical bend, as in Fig. P3.43. The total length of pipe betwee
Nataliya [291]

Answer:

F = 749 [N]

Explanation:

We must give full information on this problem, as well as the question that needs to be resolved.

Water at 20°C flows through a 5-cm-diameter pipe that has a 180° vertical bend. The total length of pipe between flanges 1 and 2 is 75 cm. When the weight flow rate is 230 N/s,

P1=165kPa

and

P2=134kPa

Neglecting pipe weight, determine the total force that the flanges must withstand for this flow.

We must make a U-shaped body diagram of the pipe in order to visualize the forces acting according to the pressures and the area of the pipe.

Then by means of the second law of motion of Newton, which says that the sum of the forces must be equal to the product of mass times acceleration, we can find the force Fp

Let us remember that pressure is defined as the divided force over the area, therefore:

F = P *A

where:

P = pressure

A = area

The product of the mass by acceleration, is equal to the product of the speed of the fluid by the mass flow, since we know the weight of the fluid we can find its mass flow.

W_{flow}=230[N/s]\\W_{flow} =g*m_{flow}\\m_{flow} = W_{flow} / g\\m_{flow} = 230/9.81\\m_{flow}= 23.45[kg/s]

In the function of the mass flow, we can find the velocity of the fluid, as we also know the diameter of the pipe

m_{flow} = density*v*A\\where\\density = 1000[kg/m^{3}]\\ v= velocity[m/s]\\A = area [m^{2}]\\v=\frac{m_{flow}}{density*A} \\v=\frac{23.45}{1000*\frac{\pi}{4}*(5*10^{-2})^{2} } \\v= 11.94 [m/s]

We know that the atmospheric pressure is equal to:

P_{atm} = 101.325[kpa]

The value of the pipe area is calculated for a circular section

A = \frac{\pi}{4} * (0.05)^{2}\\ A = 0.00196[m^{2} ]

The resultant force is 749 [N]

The solution of the equations and the free body diagram can be seen in the attached picture.

3 0
3 years ago
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that
NISA [10]

Answer:

Explanation:

We shall consider a Gaussian surface inside the insulation in the form of curved wall of a cylinder having radius equal to 3mm and unit length , length being parallel to the axis of wire .

Charge inside the cylinder = 250 x 10⁻⁹ C .

Let E be electric field at the curved surface , perpendicular to surface .

Total electric flux coming out of curved surface

= 2π r x 1 x E

= 2 x 3.14 x 3 x 10⁻³ E

According to Gauss's theorem , total flux coming out

= charge inside / ε ( 250 x 10⁻⁹C  charge will lie inside cylinder )

= 250 x 10⁻⁹ / 2.5 x 8.85 x 10⁻¹²    (  ε = 2.5 ε₀ = 2.5 x 8.85 x 10⁻¹² )

= 11.3 x 10³ weber .

so ,

2 x 3.14 x 3 x 10⁻³ E = 11.3 x 10³

E =  11.3 x 10³ /  2 x 3.14 x 3 x 10⁻³

= .599 x 10⁶ N /C .

4 0
4 years ago
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