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3 years ago

What is the length of 0.005 mm in kilometers?

1 answer:

4 0

The answer would be $5*10^9
$ .

If we divide $\frac{5*10^9}{1*10^6}$ , it gives us 0.005!

Hope this helped! c:

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As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal sprin

shepuryov [24]

Answer:

The magnitude of force must you apply to hold the platform in this position = 888.89 N

Explanation:

Given that :

Workdone (W) = 80.0 J

length x = 0.180 m

The equation for this work done by the spring is expressed as:

$W = \frac{1}{2}k_{eq}x^2$

Making the spring constant $k_{eq}$ the subject of the formula; we have:

$k_{eq} = \frac{2W}{x^2}$

Substituting our given values, we have:

$k_{eq} = \frac{2*80}{0.180^2}$

$k_{eq} = 4938.27 \ N.m^{-1}$

The magnitude of the force that must be apply to the hold platform in this position is given by the formula :

$F = k_{eq}x$

$F = 4938.27*0.180$

F = 888.89 N

6 0

3 years ago

tony walks at an average speed of 70m/min from home to school. If the difference between his home and the school is 2100 m, how

finlep [7]

Answer. 30 minutes
Explanation. If he walks 70 m in one minute how long will it take him to walk 2,100 m. Well, this is a simple division problem (you could also use a ratio box).
2100/70= 30. Hope this helps, let me know if it’s correct so others can use it :)
Good luck.

5 0

3 years ago

A young man exerted a force of 9000N on a stalled car but was unable to move it. How much work was done?

rodikova [14]

Answer:

No work was done at all.

Explanation:

Work is only done when a force is exerted and it moves an object. However, when the man exerted his force on the car, the car did not move at all. Therefore, no work was done.

7 0

3 years ago

In an incompressible three-dimensional flow field, the velocity components are given by u = ax + byz; υ = cy + dxz. Determine th

Ksivusya [100]

An incompressible flow field F in a 3D cartesian grid with components u,v,w:

F = u + v + w

where u,v,w are functions of x,y,z

Must satisfy:

∇·F = du/dx + dv/dy + dw/dz = 0

We have a field F defined:

F = u+v+w, u = ax+byz, v = cy+dxz

du/dx = a, dv/dy = c

Recall ∇·F = 0:

∇·F = du/dx + dv/dy + dw/dz = 0

a + c + dw/dz = 0

dw/dz = -a-c

Solve for w by separation of variables:

w = ∫(-a-c)dz

w = -az - cz + f(x,y)

f(x,y) is some undetermined function of x and y

The question states that w is not a function of x and y, therefore f(x,y) = 0...

w = -az - cz

8 0

3 years ago

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal

PilotLPTM [1.2K]

Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
X: mgsin65 - F = mAx 
Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) 
Moment about center of mass: 
Fr = Iα 
Now Ax = rα 
and F = umgcos65 
mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) 
umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) 
ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) 
ugcos65 = 0.4(gsin65 - ugcos65) 
1.4ugcos65 = 0.4gsin65 
1.4ucos65 = 0.4sin65 
u = 0.4sin65/1.4cos65 
u = 0.613

3 0

3 years ago