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2 years ago

14

a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi

ne the spring constant of the gun

1 answer:

Vikki [24]2 years ago

7 0

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

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A 0.110 kg cube of ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.70

Sonbull [250]

Answer:

the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

Explanation:

The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:

$V = \frac{m}{ \rho}$

Given that ;

the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g

density of the glycerine ( $\rho$ ) = 1.260 kg/L = 1.260 g/cm³

Then:

V = $\frac{0.110*10^3 \ g}{1.260 \ g/cm^3}$

V = $0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})$

V = 0.0873 L

Now;Initially the volume of the glycerin before the ice cube starts to melt is:

$V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L$

However; the volume of the water produced by the 0.11 kg ice cube = $0.11*10^3 \ cm^3$

The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

$V_2 = V_i + V"$

replacing $V"$ with $0.11*10^3 \ cm^3$ ; we have:

$V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})$

$V_2 = V_i + 0.11 \ L$

The overall total change in the volume of the glycerin is illustrated as:

$V_f = V_2 - V_1$

Now; from the foregoing ; lets replace the respective value of $V_2$ and $V_1$ in the above equation ; we have;

$V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L$

The formula usually known to be the volume of a cylinder is :

$V = \pi r ^2 h$

For the question ; we will have:

$V_f = \pi r ^2 h$

making h the subject of the formula ; we have:

$h = \frac{V_f}{\pi r^2}$

replacing 0.0227 L for $V_f$ and the given value of radius which is = 3.70 cm; we have:

$h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}$

$h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm$

Thus ; the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

8 0

2 years ago

A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i

OLga [1]

Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

7 0

2 years ago

Which of the following is not a factor in whether a reaction will spontaneously occur? A. Entropy change of the system B. Enthal

Delvig [45]

Answer:

D

Explanation:

pressure change have nothing to do with the spontaneity.

Entropy change , enthalpy change , temperature have roles in deciding spontaneity.

6 0

2 years ago

A typical garden hose has an inner diameter of 5/8". Let's say you connect it to a faucet and the water comes out of the hose wi

castortr0y [4]

Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
2) 0.25in × 1ft/12in = 0.021ft
Equation:
$v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \: \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142$
$we \: can \: only \: assume \:that \\ flow \: (q) \:stays \: same \: since \: it \\ isnt \: impeded \: by \: anything \\ thus \:it \: (q)\: stays \: the \: same \: \\ so \: 4q \: can \: be \: removed \: from \: \\ the \: equation$
$then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to > > \\ (v1 \times {d1}^{2} \pi) = (v2 \times {d2}^{2}\pi)$
$both \: \pi \: will \: cancel \: each \: other \: out \: \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \: =$

$(v1 \times {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times {0.052}^{2}) = (v2 \times {0.021}^{2}) \\ divide \: both \: sides \: by \: {0.021}^{2} \\ to \: solve \: for \: v2 > >$
$v2 = (7.0)( {0.052}^{2} ) \div ( {0.021}^{2}) \\ v2 = (7.0)(.0027) \div (.00043) \\ v2 = 44 \: feet \: per \: second$
new velocity coming out of the hose then is
44 ft/sec

4 0

2 years ago

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

a.

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

3 0

2 years ago