Answer:
The position of the particle is -2.34 m.
Explanation:
Hi there!
The equation of position of a particle moving in a straight line with constant acceleration is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the particle at a time t:
x0 = initial position.
v0 = initial velocity.
t = time
a = acceleration
We have the following information:
x0 = 0.270 m
v0 = 0.140 m/s
a = -0.320 m/s²
t = 4.50 s (In the question, where it says "4.50 m/s^2" it should say "4.50 s". I have looked on the web and have confirmed it).
Then, we have all the needed data to calculate the position of the particle:
x = x0 + v0 · t + 1/2 · a · t²
x = 0.270 m + 0.140 m/s · 4.50 s - 1/2 · 0.320 m/s² · (4.50 s)²
x = -2.34 m
The position of the particle is -2.34 m.
Answer:
In my opinion the unstoppable object will hit the unmovable object and stop but the wheels will still be rolling and trying to move but can't.
<h3>Hope this helps.</h3><h3>Good luck ✅.</h3>
When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.
<h3>Compare and contrast the energy transfer of a roller coaster to that of a pendulum:</h3><h3>What is the transfer of energy in a roller coaster?</h3>
The transfer of potential energy to kinetic energy occur when the roller coaster move along the track. As the motor pulls the cars to the top, the body has more potential energy whereas when the body comes to the bottom , it has kinetic energy in the object.
<h3>What is the energy transfer in a pendulum?</h3>
As a pendulum swings, its potential energy changes to kinetic energy and kinetic energy changes into potential energy. At the top more potential energy is present.
So we can conclude that When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.
Learn more about energy here: brainly.com/question/13881533
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Answer:
Explanation:
The voltage of a disconnected charged capacitor increases when the plate area is decreased.
When plate area decreases , capacitance C decreases , but charge Q remains constant .
Q = C V where C is capacitance and V is voltage .
when C decreases , V increases for keeping Q constant .
So the statement is true.
The electric field is dependent on the charge density on the plates.
This statement is true .
The voltage of a connected charged capacitor remains the same when the plate area is decreased .
For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .
So the statement is true .
Answer:
a
b
Explanation:
From the question we are told that
The pressure of the water in the pipe is 
The speed of the water is 
The original area of the pipe is 
The new area of the pipe is ![A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cpi%20%2A%20%20%5Cfrac%7B%5B%5Cfrac%7Bd%7D%7B2%7D%20%5D%5E2%7D%7B4%7D%20%20%3D%20%20%5Cpi%20%2A%20%20%5Cfrac%7B%5Cfrac%7Bd%5E2%7D%7B4%7D%20%7D%7B4%7D%20%3D%20%5Cpi%20%5Cfrac%7Bd%5E2%7D%7B16%7D)
Generally the continuity equation is mathematically represented as
Here 
is the new velocity
So
=> 
=> 
=> 
=>
Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure
This is mathematically represented as
Here 
is the density of water with value 
![P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20P_1%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Crho%20%5B%20v_1%5E2%20-%20v_2%5E2%20%5D)
=> ![P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20110%20%2A10%5E%7B3%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%201000%20%2A%20%20%5B%201.4%20%5E2%20-%205.6%20%5E2%20%5D)
=>
