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katrin2010 [14]

3 years ago

14

If it requires 5.0 j of work to stretch a particular spring by 1.8 cm from its equilibrium length, how much more work will be re

quired to stretch it an additional 3.7 cm ?

1 answer:

ANEK [815]3 years ago

5 0

15.277.. j. I did the problem using a proportion. an additional 3.7m to the current 1.8 cm=5.5cm.

Therefore, 5.0 j/1.8cm=x/5.5cm

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Differentiate between angular displacement and linear displacement.

Sauron [17]

Answer:

The angular displacement is not a length (not measured in meters or feet), so an angular displacement is different than a linear displacement. ... As the object rotates through the angular displacement phi, the point on the edge of the disk moves distance sa along a circular path.

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What is the chemical formula for ammonium hydroxide

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NH4OH is the answer. Hope this helps you.

8 0

3 years ago

A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t

White raven [17]

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

$\frac{v_o}{r_o} = \frac{v_1}{r_1}$

v1 = $\frac{r_1}{r_o} \ \ v_o$

let's calculate

v₁ = $\frac{1.50}{1.00} \ \ 25.0$

v₁ = 37.5 cm / s

4 0

3 years ago

: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

5 0

3 years ago

A 0.105- kg hockey puck moving at 24 m/s is caught and held by a 75-kg goalie at rest. With what speed does the goalie slide on

Veseljchak [2.6K]

Answer:

<u><em>0.03 m/s</em></u>

Explanation:

<em>Applying law of conservation of momentum, </em>

<em>m₁v₁ + m₂v₂ = (m₁ + m₂)v</em>
<em>0.105(24) + 75(0) = (0.105 + 75)v</em>
<em>75.105v = 2.52</em>
<em>v = 2.52/75.105</em>
<em>v = </em><u><em>0.03 m/s</em></u>

3 0

2 years ago