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Montano1993 [528]
3 years ago
7

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

ce d. When the block is released, it slides across a surface that is frictionless except for a rough patch of width 5.0 cm that has a coefficient of kinetic friction = 0.44. Find d, such that the block's speed after crossing the rough patch is 2.3 m/s.
Physics
1 answer:
mariarad [96]3 years ago
5 0

Answer:

d = 9.69 cm

Explanation:

given,

mass of the block = 1.2 Kg

spring force constant(k) = 730 N/m

spring is compressed = d = ?

rough patch width = 5 cm

μ_k = 0.44

work done by friction = energy lost

\mu_k mg \times x = \dfrac{1}{2}kd^2-\dfrac{1}{2}mv^2

0.44\times 1.2 \times 9.8 \times 0.05 = \dfrac{1}{2}\times 730 \times d^2-\dfrac{1}{2}\times 1.2 \times 2.3^2

3.432 = 365 d^2

d = \dfrac{3.432}{365}

d = 0.0969 m

d = 9.69 cm

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The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;
Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

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Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

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Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

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32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0
2 years ago
A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time
saul85 [17]

Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

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Let's find the time it takes to get to the river

         y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

        0 = y₀ + 0 - ½ g t²

        t = \sqrt{\frac{2y_o}{g}  }

        t = ra 2 2.5 / 9.8

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the distance traveled is

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3 years ago
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7 0
2 years ago
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Astronomers suspect that a galaxy’s type can be affected both by the conditions in the protogalactic cloud from which it forms (
Katen [24]

Answer:

The items here are describing either a condition in a later interacton or a protogalactic cloud.  The results matching with spiral and elliptical galaxy are:

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Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.

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lilavasa [31]

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