★ 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..!
Static friction is what you are looking for.
Kinetic friction is the force exerted on an already moving object, slowing it down.
Answer:
E/4
Explanation:
The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
Where;
E is the electric field
σ is the surface charge density
ε₀ is the electric constant.
Formula to calculate σ is;
σ = Q/A
Where;
Q is the total charge of the sheet
A is the sheet's area.
We are told the elastic sheet is a square with a side length as d, thus ;
A = d²
So;
σ = Q/d²
Putting Q/d² for σ in the electric field equation to obtain;
E = Q/(2ε₀d²)
Now, we can see that E is inversely proportional to the square of d i.e.
E ∝ 1/d²
The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.
From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;
E_new = E/4
Explanation:
There's not enough information in the problem to solve it. We need to know either the initial speed of the lorry, or the time it takes to stop.
For example, if we assume the initial speed of the lorry is 25 m/s, then we can find the rate of deceleration:
v² = v₀² + 2aΔx
(0 m/s)² = (25 m/s)² + 2a (50 m)
a = -6.25 m/s²
We can then use Newton's second law to find the force:
F = ma
F = (7520 kg) (-6.25 m/s²)
F = -47000 N
Answer:
1500 m/s
Explanation:
Recall that for a wave,
Speed = frequency x wavelength
here we are given frequency = 500 Hz and wavelength = 3m
simply substitute into above equation
Speed = 500 Hz x 3m
= 1500 m/s
