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1 year ago

Come si compongono due forze che agiscono in diversi punti di un corpo rigido? Oof

Physics

1 answer:

bagirrra123 [75]1 year ago

7 0

Answer:

Explanation:

I dont know if this will help but A two force member is a body that has forces (and only forces, no moments) acting on it in only two locations. In order to have a two force member in static equilibrium, the net force at each location must be equal, opposite, and collinear.

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A body A of mass 1.5kg, travelling along the positive x-axis with speed 4.5m/s, collides with another body B of mass 3.2kg which

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Momentum before the collision
x-direction:
p = m₁v₁ = 1.5 * 4.5 = 6.75
x-direction:
p = 0

momentum after the collision is conserved:
x-direction:
p = 6.75 = m₁v₁ + m₂v₂ = 1.5 * 2. 1* cos -30° + 3.2 * v₂*cos θ
y-direction:
p = 0 = m₁v₁ + m₂v₂ = 1.5 * 2.1 * sin -30° + 3.2 * v₂ * sin θ

Solve the two equations for v₂ and θ.

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2 years ago

A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvat

Stolb23 [73]

Answer: f=150cm in water and f=60cm in air.

Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:

$\frac{1}{f}$ = (nglass - ni)( $\frac{1}{R1}$ - $\frac{1}{R2}$ ).

nglass is the index of refraction of the glass;

ni is the index of refraction of the medium you want, water in this case;

R1 is the curvature through which light enters the lens;

R2 is the curvature of the surface which it exits the lens;

Substituting and calculating for water (nwater = 1.3):

$\frac{1}{f}$ = (1.5 - 1.3)( $\frac{1}{10}$ - $\frac{1}{15}$ )

$\frac{1}{f}$ = 0.2( $\frac{1}{30}$ )

f = $\frac{30}{0.2}$ = 150

For air (nair = 1):

$\frac{1}{f}$ = (1.5 - 1)( $\frac{1}{10}$ - $\frac{1}{15}$ )

f = $\frac{30}{0.5}$ = 60

In water, the focal length of the lens is f = 150cm.

In air, f = 60cm.

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2 years ago

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Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.

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2 years ago

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A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences

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Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, $q=3\times 10^{-6}\ C$