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Ipatiy [6.2K]
3 years ago
6

Help me please i need help quicklydoes the lqmp act as a resistor​

Physics
1 answer:
gayaneshka [121]3 years ago
7 0

Answer:

I don't know. sry

Explanation:

I'm really sorry

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It has thermal energy
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Light undergoes refraction as it travels from water to glass. The refractive index of water = 1.33 and refractive index of glass
cestrela7 [59]

Answer: 28°

Explanation:

Snell's law  n₁sinθ₁ = n₂sinθ₂

1.33sin32 = 1.5sinθ₂

θ₂ =28°

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3 years ago
1. Which area should one seek if caught outside in a thunderstorm?
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1. sturdy shelter

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4 years ago
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Un cuerpo se desplaza desde el punto 5,7 m al 7,5 en linea recta. cual es el valor del desplazamiento de la coordenada x?
adoni [48]

Answer:

Δx = 2m

Explanation:

From the given information:

De la información dada:

El desplazamiento ocurrió en un vector que se mueve desde una posición inicial particular hasta el final.

Entonces, supongamos que el desplazamiento es D;

Luego:

D = (7,5 ) - (5,7)

D = (2, -2)

Por lo tanto, el valor del desplazamiento de la coordenada x; Δx = 2m

8 0
3 years ago
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they
Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0
3 years ago
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