What occurs when a swimmer pushes through the water to swim?

2 answers:

6 0

When a swimmer pushes threw water to swim they are propelled forward because of the water resistance against the hand and feet.

ehidna [41]2 years ago

4 0

The correct answer to the question is that water will exert a force on the swimmer in forward direction, and swimmer will move along that direction.

EXPLANATION:

Before going to answer this question, first we have to understand Newton's third laws of motion.

Newton's third laws of motion states that, for every action, there is an equal and opposite reaction.

In the given question, the swimmer pushes the water in backward direction. This is the force of action. As per Newton's third law, the water must exert a reaction on the swimmer. The water applies a force on the swimmer in forward direction for which the swimmer can swim in the water.

You might be interested in

. Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connecte

Umnica [9.8K]

Given :

Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connected with wires as shown in the circuit given in figure. The current flowing in both the circuits is the same.

To Find :

Will the heat produced in both the cases be equal.

Solution :

Heat released is given by :

H = i²Rt

Here, R is resistance and is given by :

$R = \dfrac{\rho L}{A}$

So,

$H = i^2\times \dfrac{\rho L}{A} t\\\\H = \dfrac{i^2\rho Lt}{A}$

Now, in the question every thing is constant except for the length of the wire and from above equation heat is directly proportional to the length of the wire.

So, heat produced by Reem's wire is more than Nain one.

Hence, this is the required solution.

7 0

2 years ago

The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 μT. How far is point P from t

Tanzania [10]

Answer:

r = 0.05 m = 5 cm

Explanation:

Applying ampere's law to the wire, we get:

$B = \frac{\mu_oI}{2\pi r}\\\\r = \frac{\mu_oI}{2\pi B}$