Calculate the force at sea level that a boy of mass 50 kg exerts on a chair in which he is sitting.

Physics

2 answers:

NISA [10]3 years ago

8 0

490 N down. i just answered it well taking the quiz.

kogti [31]3 years ago

6 0

As per the question the mass of the boy is 50 kg.

The boy sits on a chair.

We are asked to calculate the force exerted by the boy on the chair at sea level.

The force exerted by boy on the chair while sitting on it is nothing else except the force of gravity of earth i.e the weight of the body .The direction of that force is vertically downward.

At sea level the acceleration due to gravity g = 9.8 m/s^2

Hence the weight of the boy $w = mg$ [m is the mass of the body]

we have m = 50 kg.

Hence w = 50 kg ×9.8 m/s^2

=490 N kg m/s^2

= 490 N

Here newton [N] is the unit of force.

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If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f

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1. The magnitude of the magnetic field doubles

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

where $\mu_0$ is the vacuum permeability, I is the current in the wire, r is the distance from the wire.

As we see from the formula, the intensity of the magnetic field is directly proportional to the current: if the current increases from 5 A to 10 A, it means it doubles, so the magnetic field doubles as well.

2. The magnitude of the magnetic field halves

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

We see that the magnitude of the magnetic field is inversely proportional to the distance from the wire (r). In this case, the distance of the particle is changed from 10 cm to 20 cm, so it is doubled: therefore, the magnitude of the field will become half of the initial value.

3. The force reverses direction

Explanation: the force exerted on a charged particle in a magnetic field is:

$F=qvB sin \theta$

where q is the charge, v is the speed of the particle, B is the magnetic field intensity and $\theta$ the angle between the direction of v and B. If the charge of the particle is switched from 2 µC to –2µC, the magnitude of the force does not change (because the absolute value of q does not change), however the charge q gets a negative sign (-), so the sign of the force changes and gets a negative sign too, so the force reverses direction.