Answer:
continuing the same eating patterns.
Explanation:
since it will only make the suituition go on for longer.
Answer:
Minimum coefficient of kinetic friction between the surface and the block is 
.
Explanation:
Given:
Mass of the block = M
Spring constant = k
Distance pulled = x
According to the question:
<em>We have to find the minimum co-efficient of kinetic friction between the surface and the block that will prevent the block from returning to its equilibrium with non-zero speed. </em>
So,
From the FBD we can say that:
⇒ Normal force, 
<em>...equation(i)</em>
⇒ Elastic potential energy, 
= 
<em> ...equation (ii)</em>
⇒ Frictional force, 
= 
<em> ...equation (iii)</em>
⇒ Plugging (i) in (iii).
⇒ 
Now,
⇒ As we know that the energy lost due to friction is equivalent to PE .
⇒ 
<em>...considering PE as</em> 
or 
.
Arranging the equation.
⇒ 
⇒ 
<em>...eliminating x from both sides.</em>
⇒ 
<em>...dividing both sides wit Mg.</em>
Minimum coefficient of kinetic friction between the surface and the block is .
An atom gains one or more electrons and becomes negatively charged, or loses one or more electrons and becomes positively charged.
The answer is to this question D
Answer:
Explanation:
Due to changing magnetic field , there will be emf induced in the region . EMF induced will create electric field which will be circular in shape and will be uniform along its circular path
The magnitude of circular electric field can be calculated as follows
We should apply Faraday law of electro magnetic induction
e = - dФ / dt = - ∫ E dl
Here Ф = π r² B
π r² dB / dt = - ∫ E dl
π r² dB / dt = E x 2π r
E = - r / 2 x dB / dt
For a circular electric field having a particular radius , magnitude of field will be constant .
The direction of electric field will be known by lenz's law
In the given case , magnetic field is upwards and it is reducing , therefore electric field induced will be such as to prevent this change of flux.
So electric field will be anticlock-wise . Hence direction of acceleration will also be anticlock-wise on proton at 1.5 cm from the centre.
