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otez555
1 month ago
10

The density of water is 1 g/cc. Ice floats on water with 90% of its volume

Physics
2 answers:
Sav [3.1K]1 month ago
8 0

Respuesta:

Explicación:

Es bastante sencillo calcularlo al darse cuenta de que la masa del agua desplazada es igual a la masa del objeto que flota en ella.

Tomemos un iceberg de volumen V. Tiene una densidad d_i. Por lo tanto, su masa total es V*d_i = m.

Ahora, está desplazando el 90% de su volumen. El agua del mar tiene una d_w. La masa total del agua desplazada es 0.9 * V * d_w, y es igual a la masa del iceberg, V*d_i.

Dividiendo ambos lados por V. Obtienes 0.9 * d_w = d_i. Ahora solo tenemos que resolver para d_w. Dividiendo ambos lados por 0.9, obtienes d_w = d_i/0.9.

Sav [3.1K]1 month ago
3 0

Respuesta: 24.3 g

Explicación:

De acuerdo con el principio de Arquímedes,

la fuerza de flotación = peso del líquido desplazado

La densidad del objeto que flota en un líquido = densidad del líquido * fracción del objeto dentro del líquido

Matemáticamente representado así:

ρ(objeto) = ρ(agua) * fracción f

ρ(agua) = 1 g/cc, fracción del objeto dentro del líquido  = 10% = 0.1

ρ(objeto) = 1 × ( 1 — 0.1 ) = 0.9 g/cc

ρ(objeto) = 0.9 g/cc

Para calcular la masa de un cubo de hielo, utilizamos la fórmula M = ρV

ρ = 0.9 g/cc, L = 3cm, V = L^3 = 3^3 = 27 cc

M = 0.9 * 27

M = 24.3 g

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The amount of work performed by a system at consistent pressure is defined by the following equation:
W=p \Delta V = p (V_f - V_i)
where
p represents pressure
V_f as the final volume
V_i as the initial volume

Plugging the values given in this case into the formula gives us
W=p (V_f -V_i)=(1.00 atm)(2.00 L-1.00 L)=1.00 L\cdot atm

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2 months ago
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A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from w
ValentinkaMS [3465]

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at   ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

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The horizontal distance covered is

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8 0
2 months ago
A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu
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Respuesta:

La magnitud de la aceleración resultante es 2.2 m/s^2

Explicación:

La masa (m) del velero es 2000 kg

La fuerza que actúa sobre el velero debido a la marea del océano es F_1 = 3000N

Hacia el este significa que se da a lo largo de la dirección positiva del eje x

EntoncesF_{1x} = 3000N y F_{1y}= 0

La fuerza del viento que actúa sobre el velero esF_2 6000N dirigida hacia el noroeste, lo que significa a un ángulo de 45 grados sobre el eje negativo x

Luego

F_{2x} = -(6000N) cos 45 grados = -4242.6 N

F_{2y} = (6000N) cos 45 grados = 4242.6 N

Por lo tanto, la fuerza neta que actúa sobre el velero en la dirección x es

F_x = F_{1x}+ F_{2x}

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

La fuerza neta que actúa sobre el velero en la dirección y es

= 0+ 4242.6N

= 4242.6N F_y = F_{1y}+ F_{2y}La magnitud de la fuerza resultante =

Usando el teorema de Pitágoras de 1243 N y 4243 N

4420.8 N\sqrt{(1242.6)^2 + (4242.6)^2F = ma

\sqrt{(1544054.76) + (17999654.8)}

\sqrt{(19543709.5)^2}

= 2.2

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