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3 years ago

If a suspended object A is attracted to a charged object B, can we conclude that A is charged?

1 answer:

8 0

Not necessarily, object A could also be neutral, and becoming a dipole due to object B's charge. A charged object can induce a dipole in a neutral object, and that object would then become attracted without being charged.

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The bodies in this universe attract one another name the scientist who propounded this statement

Colt1911 [192]

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7 0

3 years ago

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

3 years ago

A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio

jeka57 [31]

Answer:

a. $3.1 m/s^2$

Explanation:

The equation of the forces along the directions parallel and perpendicular to the slope are:

- Along the parallel direction:

$mg sin \theta - \mu_k R = ma$

where :

m = 6.0 kg is the mass of the box

g = 9.8 m/s^2 the acceleration of gravity

$\theta=39^{\circ}$ is the angle of the slope

$\mu_k = 0.40$ is the coefficient of friction

R is the normal reaction

a is the acceleration

- Along the perpendicular direction:

$R-mg cos \theta =0$

From the 2nd equation, we get an expression for the reaction force:

$R=mg cos \theta$

And substituting into the 1st equation, we can find the acceleration:

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta =(9.8)(sin 39^{\circ})-(0.40)(9.8)(cos 39^{\circ})=3.1 m/s^2$

6 0

3 years ago

Ann is driving down a street at 61 km/h.

Wittaler [7]

Answer:

12.267 seconds approximately.

Explanation:

The units can be simplified into m/s, in which case you would have 61000/3600. Simplify that to 16 and 17/18. This is your meters per second, so multiply that by .724 to get the answer.

4 0

4 years ago

Lightweight, vertically suspended spiral spring with a spring constant of 8.6 N / m is fitted with 64 g weight. The weight shall

alexandr1967 [171]

Answer:

Explanation:

Given that

Force constant k=8.6N/m

Weight =64g=64/1000=0.064kg

Extension is 45mm=45/1000= 0.045m

It will have it highest spend when the Potential energy is zero

Therefore energy in spring =change in kinetic energy

Ux=∆K.e

½ke² = ½mVf² — ½mVi²

Initial velocity is 0, Vi=0m/s

½ke² = ½mVf²

½ ×8.6 × 0.045² = ½ ×0.064 ×Vf²

0.0087075 = 0.032 Vf²

Then, Vf² = 0.0087075/0.032

Vf² = 0.2721

Vf=√0.2721

Vf= 0.522m/s

The time it will have this maximum velocity?

Using equation of motion

Vf= Vi + gr

0.522= 0+9.81t

t=0.522/9.81

t= 0.0532sec

t= 53.2 milliseconds

5 0

3 years ago