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kvasek [131]
3 years ago
14

There are two blocks: one large one initially at rest, and a smaller one, initially moving to the right withsome speed. The smal

l block is 25 kg, and the large block is 50 kg. The blocks are to the left of a hill.The hill is 10 meters tall. (Drawing not to scale)a) The small block collides with the large block and sticks together. How fast must the initial velocityof the small block be so that the two blocks just reach the top of the hill. (Assume no friction).b) What was the total impulse the small block exerted on the large block in order to get it going fastenough
Physics
1 answer:
KATRIN_1 [288]3 years ago
6 0

Answer:

Explanation:

Let the initial velocity of small block be v .

by applying conservation of momentum we can find velocity of common mass

25 v = 75 V , V is velocity of common mass after collision.

V = v / 3

For reaching the height we shall apply conservation of mechanical energy

1/2 m v² = mgh

1/2  x 75 x V² = 75 x g x 10

V² = 2g x 10

v² / 9 = 2 x 9.8 x 10

v² = 9 x 2 x 9.8 x 10

v = 42 m /s

small block must have velocity of 42 m /s .

Impulse by small block on large block

= change in momentum of large block

= 75 x V

= 75  x 42 / 3

= 1050 Ns.

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In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 2.8 m/s. However, this speed is inade
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3.95979 m/s

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u = Initial velocity

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v^2-u^2=2as\\\Rightarrow as=\frac{v^2-u^2}{2}\\\Rightarrow as=\frac{0^2-2.8^2}{2}\\\Rightarrow as=-3.92

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8 0
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A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide suffi
Mademuasel [1]

The image of the water tower and the houses is in the attachment.

Answer: (a) P = 245kPa;

(b) P = 173.5 kPa

Explanation: <u>Gauge</u> <u>pressure</u> is the pressure relative to the atmospheric pressure and it is only dependent of the height of the liquid in the container.

The pressure is calculated as: P = hρg

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ρ is the density of the liquid, in this case, water, which is ρ = 1000kg/m³;

When it is full the reservoir contains 5.25×10⁵ kg. So, knowing the density, you know the volume:

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To know the height of the spherical reservoir, its diameter is needed and to determine it, find the radius:

V = \frac{4}{3}.\pi.r^{3}

r = \sqrt[3]{ \frac{3}{4\pi } .V}

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h = 17.71

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P = 17.71.1000.9.8

P = 173.5.10³ Pa or 173.5 kPa

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3 years ago
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