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zloy xaker [14]
3 years ago
11

A 71-kg swimmer dives horizontally off a 500-kg raft. If the diver's speed immediately after leaving the raft is 6m/s, what is t

he corresponding raft speed?
Physics
1 answer:
Sloan [31]3 years ago
5 0

Answer:

The answer is below

Explanation:

Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.

Momentum = mass * velocity

The principle of conservation of momentum states that momentum cannot be created or destroyed but  can be transferred. Therefore the momentum before and after an action is equal.

Initial momentum = Final momentum

Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.

m = 71 kg, M = 500 kg, v = 6 m/s

Initial both the raft and diver are at rest, hence u and U is zero, hence:

mu + MU = mv + MV

71(0) + 500(0) = 71(6) + 500(V)

0 = 426 + 500(V)

500(V) = -426

V = -426/500

V = -0.852 m/s

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The complete question is :

A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?

4 0
2 years ago
Two positive charges are fixed a distance apart.the sun of their charges is Qt.what charge must each have in order to maximise t
romanna [79]

Answer:

Both charges must have the same charge, Qt/2.

Explanation:

Let the two charges have charge Q1 and Q2, respectively.

Use Coulombs's Law to find an expression for the force between the two charges.

F = k_e\frac{Q_1Q_2}{r^2}, where

Ke is Coulomb's contant and

r is the distance between the charges.

We know from the question that

Q1 + Q2 = Qt

So,

Q2 = Qt - Q1

F = k_e\frac{Q_1(Q_t - Q_1)}{r^2}

Simplify to obtain,

F = \frac{k_e}{r^2} (Q_tQ_1 - Q_1^2)

In order to find the value of Q1 for which F is the maximum, we will use the optimization technique of calculus.

Differentiate F with respect to Q1,

\frac{dF}{dQ_1}  = \frac{k_e}{r^2} (Q_t - 2Q_1)

Equate the differential to 0, to obtain the value of Q1 for which F is the maximum.

\frac{k_e}{r^2} (Q_t - 2Q_1) = 0\\Q_t - 2Q_1 = 0\\2Q_1 = Q_t\\Q1 = \frac{Q_t}{2}

It follows that

Q_2 = \frac{Q_t}{2}.

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