All categories

3 years ago

Greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but

Physics

1 answer:

maw [93]3 years ago

5 0

${\large{\bold{\rm{\underline{Given \; that}}}}}$

★ 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.

${\large{\bold{\rm{\underline{To\; find}}}}}$

★ The speed of the hound and the hare

${\large{\bold{\rm{\underline{Solution}}}}}$

★ The speed of the hound and the hare = 25:18

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

$\dashrightarrow$ 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..!

(5/3)a × 5

25/3a metres

Now the distance travelled by the hare in it's 6 leaps..!

6a metres

Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

25/3a = 6a

25/3 = 6

25:18

You might be interested in

Kyle has a mass of 54kg and is moving at 3 m/s what is his kinetic energy

Mars2501 [29]

Answer:

243J

Explanation:

K.E = 1/2 x 54 x 3^2

K.E = 1/2 x 54 x 9

K.E = 1/2 x 486

K.E = 486/2

K.E = 243J

6 0

3 years ago

If the pendulum took longer to complete one oscillation, how would the graph change?

Sindrei [870]

Answer:

took longer to complete one oscillation, that means its PERIOD increased, and the distance between the peaks of the graph would be longer.

line would be less. the period of oscillation would have any effect on the graph

7 0

3 years ago

Read 2 more answers

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$