If an aqueous solution has a slippery feel, which type of solution will it most likely be

A force of F = (2.00ˆi + 3.00ˆj) N is applied to an object that is pivoted about a fixed axle aligned along the z coordinate axi

Vladimir [108]

Explanation:

It is given that,

Force applied to object, $F=(2i+3j)\ N$

Position, $r=(4i+5j)\ m$

(b) The cross product of force and position vector is used to find the net torque about the z axis. It is given by :

$\tau=F\times r$

$\tau=(2i+3j) \times (4i+5j)$

$\tau=\begin{pmatrix}0&0&-2\end{pmatrix}$

or

$\tau=(-2k)\ N-m$

The torque is acting in -z axis.

(a) The magnitude of torque is given by :

$|\tau|=\sqrt{(-2)^2}$

$|\tau|=2\ N-m$

Hence, this is the required solution.

5 0

3 years ago

Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

VARVARA [1.3K]

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

8 0

3 years ago

At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction

bogdanovich [222]

Answer:

magnitude of force on charge 2Q = $\frac{KQ^{2} }{I^{2} }$

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force

║F║ = $\sqrt{f_{x}^{2} + f_{y}^{2} }$ = after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW