Answer:
![\frac{dV}{dt}= \frac{\pi d^2}{4}v](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7Dv)
Explanation:
The rate of volume flow out of tank can be expressed as:
![\frac{dV}{dt} = A\frac{dL}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%20%3D%20A%5Cfrac%7BdL%7D%7Bdt%7D)
where,
dV/dt = Volume flow rate
A = Cross-sectional area of outlet = πd²/4
d = diameter of circular outlet
dL = Displacement covered by water
dt = time taken
but we know that:
Velocity = υ = displacement/time = dL/dt
Substituting the values of "dL/dt" and "A" in the equation, we get:
![\frac{dV}{dt} = \frac{\pi d^2}{4}v](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%20%3D%20%5Cfrac%7B%5Cpi%20d%5E2%7D%7B4%7Dv)
This is the expression for volume flow rate dV/dt, on terms pf v, d.
Answer:
a= g = - 9.81 m/s2.
The following equations will be helpful:
a = (vf - vo)/t d = vot + 1/2 at2 vf2 = vo2 + 2ad
When you substitute the specific acceleration due to gravity (g), the equations are as follows:
g = (vf - vo)/t d = vot + 1/2 gt2 vf2 = vo2 + 2gd
If the object is dropped from rest, the initial velocity ("vi") is zero. This further simplifies the equations to these:
g = vf /t d = 1/2 gt2 vf2 = 2gd
The sign convention that we will use for direction is this: "down" is the negative direction. If you are given a velocity such as -5.0 m/s, we will assume that the direction of the velocity vector is down. Also if you are told that an object falls with a velocity of 5.0 m/s, you would substitute -5.0 m/s in your equations. The sign convention would also apply to the acceleration due to gravity as shown above. The direction of the acceleration vector is down (-9.81 m/s2) because the gravitational force causing the acceleration is directed downward.
hope this info helps you out!
It can be explained as follows: consider the field produced by a positive charge. If we place a positive test charge in this a field, then this charge would move away from the central charge (because like charges repel), while if we place a negative test charge in this field, this charge would move towards the central charge (because opposite charges repel)
Explanation:
Electric fields are vector fields, and they are represented using field lines.
The field lines give indications on both the magnitude and the direction of the electric field. In fact:
- The magnitude of the field can be inferred from the spacing between the lines: the closer the lines are, the stronger the field, while for a weaker field the lines are more spread apart
- The direction of the field is given by the direction of the field lines
In particular, by convention the direction of the field lines represent the direction of the force that a positive test charge would feel when immersed in that field: this means that a positive test charge would accelerate in the direction of the field lines, while a negative test charge would accelerate in the direction opposite to the field lines.
This is in agreement with the fact that like charges repel and opposite charges attract. In fact, the lines of the electric field produced by a single-point positive charge point away from the positive charge: if we place a positive test charge in this field, then this charge would move away from the central charge (because like charges repel), while if we place a negative test charge in this field, this charge would move towards the central charge (because opposite charges repel).
Learn more about electric fields:
brainly.com/question/8960054
brainly.com/question/4273177
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Answer:
True the plastic will float because of the principle of flotation or buoyancy
Explanation:
Buoyancy explains it all!!
Buoyancy is the upward force/upthrust experienced by a body immersed totally or partially in a liquid.
According to the principle of flotation:
<em>"when a body is totally or partially immersed in liquid it experiences an upthrust which is equal to the volume of fluid displaced"</em>
The plastic will float due to the fact the average density of the total volume of the plastic and the air inside it is less than the same volume of water it is floating in