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Travka [436]
2 years ago
7

In which situation is work not being done?

Physics
1 answer:
almond37 [142]2 years ago
5 0

AS

work done =W = F.d = F d cosФ     (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body.  F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0

Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0

example:  A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.

The third term upon which work done  dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0   ( as cos 90°=0)

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inna [77]

Answer:

0.025 m

Explanation:

From the question,

Applying Hook's law

F = ke................... Equation 1

Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.

make e the subject of the equation

e = F/k....................... Equation 2

Given: F = 10 N, e = 395 N/m

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e = 10/395

e = 0.025 m

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3 years ago
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8 0
2 years ago
A 6.4-N force pulls horizontally on a 1.5-kg block that slides on a smooth horizontal surface. This block is connected by a hori
Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless.  I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
                                    Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
                                                                 </em>
That's about  <em>2.634 m/s²</em>  <em>

</em>
(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

       Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.

That's the force that's accelerating the little block, so that must be the tension
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7 0
2 years ago
A mass weighting 16 lbs stretches a spring 3 inches. The mass is in a medium that exerts a viscous resistance of 20 lbs when the
const2013 [10]

Answer:

The equation for the object's displacement is u(t)=0.583cos11.35t

Explanation:

Given:

m = 16 lb

δ = 3 in

The stiffness is:

k=\frac{m}{\delta } =\frac{16}{3} =5.33lb/in

The angular speed is:

w=\sqrt{\frac{k}{m} } =\sqrt{\frac{5.33*386.4}{16} } =11.35rad/s

The damping force is:

F_{D} =cu

Where

FD = 20 lb

u = 4 ft/s = 48 in/s

Replacing:

c=\frac{F_{D} }{u} =\frac{20}{48} =0.42lbs/in

The critical damping is equal:

c_{c} =\frac{2k}{w} =\frac{2*5.33}{11.35} =0.94lbs/in

Like cc>c the system is undamped

The equilibrium expression is:

u(t)=u(o)coswt+u'(o)sinwt\\u(o)=7=0.583\\u'(o)=0\\u(t)=0.583coswt\\u(t)=0.583cos11.35t

3 0
3 years ago
A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini
Volgvan

Answer:

The horizontal range will be 2.55\times 10^5m

Explanation:

We have given initial speed of the shell u = 1.6\times 10^3m/sec

Angle of projection = 51°

Acceleration due to gravity g=9.8m/sec^2

We have to find maximum range

Horizontal range in projectile motion is given by

R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m

So the horizontal range will be 2.55\times 10^5m

6 0
2 years ago
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