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2 years ago

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A worker pushes a 7 kg shipping box along a roller track. Assume friction is small enough to be ignored because of the rollers.

The worker's push is 25 N directed down and to the right at an angle of 20°.
(a) Determine the horizontal component of the worker's push.
(b) Write a net force equation for the horizontal forces on the box.

1 answer:

Bingel [31]2 years ago

6 0

Answer:

a) Fₓ = 23.5 N

b) Net force = Fₓ

Explanation:

An image of the question as described is attached to this solution.

From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)

a) From the diagram, the horizontal component of the force is

Fₓ = 25 cos 20° = 23.49 N = 25 N

b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain

Net force = Fₓ - Frictional force

But frictional force is 0 N

Net force = Fₓ

Hope this Helps!!!

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Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

<em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

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Genrish500 [490]

Answer:

The volume at mountains is 2.766 L.

Explanation:

Given that,

Volume $V_{1} = 2.00\ L$

Pressure $P_{1}= 1.00\ atm$

Pressure $P_{2}= 70.0\ kPa$

Temperature $T_{1}= 20.0°C = 293\ K$

Temperature $T_{2}= 7.00°C = 280\ K$

We need to calculate the volume at mountains

Using gas law

$\dfrac{PV}{T}=\ Constant$

For both temperature,

$\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$

Put the value into the formula

$\dfrac{101.325\times2}{293}=\dfrac{70\times V_{2}}{280}$

$V_{2}=\dfrac{101.325\times2\times280}{293\times70}$

$V_{2}=2.766\ litre$

Hence, The volume at mountains is 2.766 L.

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2 years ago

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Aloiza [94]

DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150

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(a) A velocity selector consists of electric and magnetic fields described by the expressions E = E and B = B ĵ, with B = 10.0 m

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Answer:

(a) 160000 kV/m

(b) 1336 keV

Explanation:

(a) magnetic filed, B = 10 T

energy of electron, E = 740 eV

mass of electron, m = 9.1 x 10^-31 kg

Let v be the velocity of electron.

E = 1/2 mv^2

740 x 1.6 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 1.6 x 10^7 m/s

v = E / B

E = v x B = 1.6 x 10^7 x 10 = 16 x 10^7 V/m

E = 160000 kV/m

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B = 10 T

Let v be the velocity of protons.

v = E / B = 16 x 10^7 / 10 = 1.6 x 10^7 m/s

Kinetic energy of proton, E = 1/2 mv^2

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sesenic [268]

Work done by a given force is given by

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here on sled two forces will do work

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2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

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$W_2 = Fdcos\theta$

$W_2 = 4*5cos180$

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Now total work done on sled

$W_{net}= W_1 + W2$

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