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

1 answer:

7 0

There is only one force applying to the boy, which is gravity in that regard pointing from the chair down to earth. The gravity force "Fg" is equal to the mass "m" of the object multiply by the earth accelaration "g". And so F = 50kg.9.8m/s = 490Newtons (N)

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In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m a

balu736 [363]

Answer:

The value is $\lambda = 214.3 \ nm$

Explanation:

From the question we are told that

The slit separation is $d = 3.00 * 10^{-5} m$

The distance of the screen is $D = 2.00\ m$

The order of fringe is n = 7

The path difference is $y = 10.0 \ cm = 0.1 \ m$

Generally the path difference is mathematically represented as

$y = \frac{n * \lambda * D}{ d}$

=> $0.1 = \frac{7 * \lambda * 2.00 }{ 3.00 * 10^{-5}}$

=> $\lambda = \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }$

=> $\lambda = 2.143 *10^{-7} \ m$

=> $\lambda = 214.3 \ nm$

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3 years ago

When a 100-N force acts horizontally on an 8.0-kg chair, the chair moves at a constant speed across the level floor. Which state

TiliK225 [7]

Answer:

The applied force is greater than the frictional force.

Explanation:

the chair moves at a constant speed therefore, the answer is not A or C.

if there is no friction then the chair would accelerate and it would not be at a constant speed.

hence, the only possible answer is B.

7 0

3 years ago

A grandfather clock depends on the period of a pendulum to keep correct time.(ii) Suppose a grandfather clock is calibrated corr

ANEK [815]

The grandfather clock will now run slow (Option A).

<h3>What is Time Period of an oscillation?</h3>

The time period of an oscillation refers to the time taken by an object to complete one oscillation.
It is the inverse of frequency of oscillation; denoted by "T".

Now,

$\sqrt{\frac{L}{g}}$ , where L is the length and g is the gravitational constant, is the formula for a pendulum's period.
The period will increase as one climbs a very tall mountain because g will slightly decrease.
Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.