Answer:
Resistivity 
It depends upon cross sectional area and length of material
Explanation:
The resistance of any material is given by 
, here 
is the resistivity of material , l is length of material and A is cross sectional area
So resistivity
So resistuivity of any material depends upon area of cross section and length of material
If cross sectional area will be more then resistivity will be more. And is length of the material will be more then resistivity will be less
Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
If the intensity of the yellow light increased, meaning more photons will strike the Potassium metal per unit area. This will cause more ejection of electrons from the metal and hence, the strength of current will also increase as we know that
I = Q/t, as the charge increase , the current will also increase.
Answer:
Explanation:
1) Hypermetropia (better known as Farsighted- this is why nearby objects seem blurry for him)
2) In such instances, image are typically formed farther from the near point
3) Such defects are quite common so there are common procedures such as using convex lens which can restore the sight to normal.
★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
★ The speed of the hound and the hare
★ The speed of the hound and the hare = 25:18
As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.
So firstly let us assume a metres as the distance covered by the hare in one leap.
Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.
But 3 leaps of the hound are equal to 5 leaps of the hare.
Henceforth, (5/3)a meters is the distance that is covered by the hound.
Now according to the question,
Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)
Now the distance travelled by the hound in it's 5 leaps..!
Now the distance travelled by the hare in it's 6 leaps..!
Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!
