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3 years ago

12

Particle a has twice the charge of nearby particle

1 answer:

bulgar [2K]3 years ago

8 0

The force between charged particles is calculated through the Coulomb's law,
F = kQ₁Q₂/d²
From the equation, it can be deduced that the force is linearly directly proportional to the charge. This means that when the charge is doubled, the force is likewise doubled. Therefore, the compared to force on b, the force on a is twice.

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The difference between a need and a want gets blurry for some people because of:

Anna [14]

The answer should be all of the above

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3 years ago

The density of a substance is 3.4 g cm-3. Its relative density relative to another substance is 2.0. what is the density of the

ankoles [38]

Answer:

1.7 g/cm³

Explanation:

Given that:

Density of substance = 3.4 g/cm³

Relative density to another substance = 2

Density of second substance=?

Let density of second substance = x

Relative density = density of substance / density of second substance

Relative density = density of substance / x

2.0 = 3.4g/cm³ / x

2 * x = 3.4 g /cm³

x = 3.4 g/cm³ ÷ 2

x = 1.7 g/cm³

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3 years ago

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cluponka [151]

They need water and sunlight to survive

8 0

3 years ago

nirvana33 [79]

#1

As per work energy theorem work done by all forces must be equal to change in its kinetic energy

so correct answer will be

$W_{net}$ = change in kinetic energy

#2

Torque is defined as

$\tau = r \times F$

$\tau = rFsin\theta$

so here torque will be maximum when applied force is perpendicular to the rod

so first figure of rod is correct for maximum torque

#3

When car makes turn then its passengers moves outwards which is due to the property of inertia because due to this property passengers has tendency to move in same direction and hence it feels to move outwards

So correct answer will be

<em>inertia</em>

#4

As per energy conservation we can say

initial potential energy = final kinetic energy

$PE = KE$

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$KE = 0.5(9.8)(0.5)$

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7 0

4 years ago

Link AB is to be made of a steel for which the ultimate normal stress is 450 MPa. Determine the cross-sectional area for AB for

ad-work [718]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The cross-sectional area is $A =1.6815 m^2$

Explanation:

The free body diagram of the link is shown on the second uploaded image

From the question we are told that

Ultimate normal stress in the link $AB = 450MPa$

Factor safety $=3.59$

From our free diagram we can see that the moment about B is 0 Mathematically

$\sum M_b =0$

But $\sum M_b = -(20*0.4)-\frac{8(1.2)^2}{20} + D_y (0.8)$

Hence $-(20*0.4)-\frac{8(1.2)^2}{20} + D_y (0.8) =0$

Making $D_y$ the subject

$D_y = 17.2kN$

At equilibrium summation of all force is 0 mathematically

This means

$\sum F_y =0$

i.e $F_{BA} sin 35^o +D_y - 8(1.2) -20 =0$

$F_{BA} = \frac{8(1.2)+20-17.2}{sin35^o}$

$F_{BA} =21.62kN$

The factor of safety is mathematically

Factor of safety $= \frac{\sigma _u}{\sigma _{ all}}$

Where $\sigma_u$ is the normal stress

$\sigma_{all}$ is the allowable stress this mathematically given as

$\sigma_{all} = \frac{F_{AB}}{A}$

$3.5 = \frac{21.62*10^3}{A}$

$Factor\ of \ safety =\frac{450*10^6}{[\frac{21*10^3}{A} ]}$

Making A the subject

$A = \frac{3.5*21*10^3}{450*10^6}$

$= 1.6815*10^{-4} m^2$

4 0

3 years ago