What is the spring constant for a supermarket scale that stretches 0.01 m when a force of 4 N is applied

Gnoma [55]

The answer is apogee my dude

7 0

3 years ago

The three components of velocity in a velocity field are given by u = Ax + By + Cz, v = Dx + Ey + Fz, and w = Gx + Hy + Jz. Dete

Alexxandr [17]

Answer:

The relationship is only between the coefficients A, E and J which is:

$A + E + J = 0$ . The remaining coefficients can be anything without any constraints.

Explanation:

Given:

The three components of velocity is a velocity field are given as:

$u = Ax + By + Cz\\\\v = Dx + Ey + Fz\\\\w = Gx + Hy + Jz$

The fluid is incompressible.

We know that, for an incompressible fluid flow, the sum of the partial derivatives of each component relative to its direction is always 0. Therefore,

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$

Now, let us find the partial derivative of each component.

$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(Ax+By+Cz)\\\\\frac{\partial u}{\partial x}=A+0+0=A\\\\\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(Dx+Ey+Fz)\\\\\frac{\partial v}{\partial y}=0+E+0=E\\\\\frac{\partial w}{\partial z}=\frac{\partial }{\partial z}(Gx+Hy+Jz)\\\\\frac{\partial w}{\partial z}=0+0+J=J$

Hence, the relationship between the coefficients is:

$A+E+J=0$

There is no such constraints on other coefficients. So, we can choose any value for the remaining coefficients B, C, D, F, G and H.

6 0

3 years ago

How long does it take a man to travel 6 km if his speed is 3km/h?

Ymorist [56]

why did my answer get deleted??

oh yeah i put a link on there- oopsies.

I wont this time!

I got 30!

3 0

2 years ago

Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin

likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0 and T - mgcos57 - F = m(0) = 0

Mg - T = 0 (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0

3 years ago

what will happen to the brightness of the other bulbs in a series circuit if any of the bulbs were to burn out?

Ivenika [448]

Answer:

not work.

Explanation:

if u are saying in a series circuit...

if 1 build burns out and theres other bulbs the circuit wont work anymore.

6 0

3 years ago