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RSB
2 months ago
10

The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2

m long is used to rotate a 0.5kg object in a circle on a frictionless table with the string horizontal. The maximum speed that the mass can attain under these conditions without the string breaking is most nearly:
Answer is 20 m/s
Please show me how I can get the answer. Thank you.
Physics
1 answer:
serg [3.5K]2 months ago
0 0

Respuesta:

Explicación:

Al analizar esta pregunta, considera el movimiento circular. Primero, determina la máxima fuerza que puede aplicarse al hilo. F = mg, entonces F = (10)(10) = 100 N. Luego, calcula la aceleración centrípeta de la masa de 0.5 kg, a = F/m, así que a = 100/.5 = 200 m/s². En la hoja de ecuaciones, usa la fórmula a (aceleración centrípeta) = v²/r, por lo que 200 = v²/2; por consiguiente, v = 20 m/s. ¡Espero que esto sea útil!

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The ballistic pendulum was invented in 1742 by English mathematician Benjamin Robins. It consists of an initially stationary pen
serg [3582]

Answer:

What is your question?

Please feel free to modify or inquire in the comments.

Thank you.

Explanation:

7 0
1 month ago
An aluminum rod is 10.0 cm long and a steel rod is 80.0 cm long when both rods are at a temperature of 15°C. Both rods have the
serg [3582]

Response:

0.9 cm

Clarification:

The following illustrates the calculation of the combined rod's length increase:

As established

Length increase = expansion of aluminum rod + expansion of steel rod

= 10cm \times 2.4e - 5\times (90-15) + 80cm\times 1.2e - 5\times (90-15)

= 0.9 cm

We simply summed the expansions of both the aluminum and steel rods to determine the overall increase in the joined rod's length, which must be factored in

4 0
1 month ago
A trained sea lion slides from rest down a long
Keith_Richards [3271]

Answer:

1.5 m/s²

Explanation:

Begin by sketching a free body diagram.  Three forces are at play on the sea lion: the force of gravity acting downwards, the normal force that is perpendicular to the ramp, and the frictional force parallel to the ramp.

Considering the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

This gives us N = mg cos θ

Next, examining the forces parallel to the incline:

∑F = ma

mg sin θ − Nμ = ma

Substituting for N yields:

mg sin θ − (mg cos θ) μ = ma

g sin θ − g cos θ μ = a

hence a = g (sin θ − μ cos θ)

If we set θ = 23° and μ = 0.26:

a = 9.8 (sin 23 − 0.26 cos 23)

this results in a = 1.48

When rounded to two significant figures, the acceleration of the sea lion is 1.5 m/s².

5 0
11 days ago
A glass tube is filled with hydrogen gas.  An electric current is passed through the tube, and the tube begins to glow a pinkish
inna [3103]
The initial description is the accurate one.
6 0
1 month ago
Read 2 more answers
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
ValentinkaMS [3465]

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

4 0
15 days ago
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