i would choose a for 945 and a for faster
Answer:
The mass of unknown object is 8.62Kg
Explanation:
To develop this problem it is necessary to apply the equations related to the Drag force and the Force of Gravity.
For the given point, that is, the moment at which the terminal velocity is reached, the two forces equalize, that is,
By definition we know that the Drag force is defined as
Where,
Drag coefficient
Density
A =Cross-sectional Area
V = Velocity
In the other hand we have,
Where,
Mass of sphere
Mass of unknown object
Equating the two equations we have to
Re-arrange for m_2,
Our values are given by,
Replacing in the equation we have,
<em>Therefore the mass of unknown object is 8.62Kg</em>
A fatter handle. This allows you to provide more torque to the screw, which should eventually loosen it.
Answer:
The charges under study are of the same sign
The calculation of the electric field for each charge separately, there is no relationship between the charges
Explanation:
Let's start by writing the equation for the electric field
E = k q / r²
where q is the charge under analysis and r the distance from this charge to a positive test charge.
When analyzing the statement the student has some problems.
* The charges under study are of the same sign, it does not matter if positive or negative.
* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.
* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero
1000 miles = 1 609 340 m
2 weeks = 1209 600 s
v = 1609340/1209600 = 1.33 m/s
