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2 years ago

3. A 75kg man sits at one end of a uniform seesaw pivoted at its center, and his 24kg son sits at the

Physics

1 answer:

bulgar [2K]2 years ago

7 0

Answer:

The wife have to sit at 0.46 L from the middle point of the seesaw.

Explanation:

We need to make a sketch of the seesaw and the loads acting over it.

And by the studying of the Newton's law we can find the equation useful to find the distance of the mother sitting on the seesaw with respect to the center ot the pivot point.

A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.

The maximum momentum with respect to the pivot point (0) will be:

$M=75 *\frac{L}{2}$

Where L/2 is the half of the distance of the seesaw

Therefore the other loads ( mom + son) must be create a momentum equal to the maximum momentum.

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alexira [117]

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2 years ago

. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

29.4 N/m

0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

m1 = 1.29 kg

m2 = 0.3 kg

x1 = -0.75 m

x2 = -0.2 m

g = 9.8 m/s^2

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

l= x1 — (F_1/k)

= 0.1

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2 years ago

A small block of wood of inertia mb is released from rest a distance h above the ground, directly above your head. you decide to

Zinaida [17]

Since I'm assuming that its perfectly elastic, considering there's not enough information given, so I think that no energy is dissipated in the collision

hmax = h - d + { [ mpvp - mb√(2gd) ] / (mp+mb) }² / (2g)

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2 years ago

A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitca

Hatshy [7]

Answer:

0.0367

Explanation:

The loss in kinetic energy results into work done by friction.

Since kinetic energy is given by

KE=0.5mv^{2}

Work done by friction is given as

W= umgd

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Making u the subject of the formula then we deduce that

$u=\frac {v^{2}}{2gd}$