Answer:
The accurate statements are
2. The train is not an inertial frame of reference.
5. The train could be moving at a constant velocity in a circular path.
8. The train must be undergoing acceleration.
Explanation:
As we observe that the string forms an angle with the horizontal
we can formulate the force equation relevant to the given ball
similarly in the Y direction
Thus we conclude
This leads us to deduce that the train is accelerating with an acceleration identical to that of gravity
The correct statements will be
2. The train is not an inertial frame of reference.
5. The train could be moving at a constant speed in a circular path.
8. The train must be experiencing acceleration.
Answer:
The pressure measured at this moment is 0.875 mPa
Explanation:
Given that,
Flow energy = 124 L/min
Boundary to system P= 108.5 kJ/min
We are tasked with finding the pressure here
Applying the pressure formula
Here, 
Where, v refers to velocity
Insert the values into the equation
Therefore, the pressure at this moment is 0.875 mPa
<span>We will apply the momentum-impulse theorem here. The total momentum along the x-direction is defined as p_(f) = p_(1) + p_(2) + p_(3) = 0.
Therefore, p_(1x) = m1v1 = 0.2 * 2 = 0.4. Additionally, p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3, where v3 represents the unknown speed and m3 signifies the mass of the third object, which has an unspecified velocity.
In the same way, for the particle of 235g, the y-component of the total momentum is described with p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
Thus, p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3, where m3 is the mass of the third piece.
Consequently, p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; yielding v3 = 0.4/-0.1 = - 4.
Similarly, p_(fy) = 0.3525 + 0.1v3; thus v3 = - 0.3525/0.1 = -3.525.
Therefore, the x-component of the speed of the third piece is v_3x = -4 and the y-component is v_3y = 3.525.
The overall speed is calculated as follows: resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
