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3 years ago

An inclined plane is used to lift a box into a moving truck. The ramp is 6m long and 1.5m

Physics

1 answer:

ElenaW [278]3 years ago

5 0

Given :

An inclined plane is used to lift a box into a moving truck. The ramp is 6 m long and 1.5 m.

To Find :

The mechanical advantage of the inclined plane.

Solution :

Mechanical advantage, M.A of an inclined plane is given by :

$M.A = \dfrac{\text{Length of slope}}{\text{Height of plane}}$

Putting given values in above equation, we get :

$M.A = \dfrac{6 \ m}{1.5 \ m}\\\\M.A = 4$

Therefore, the mechanical advantage of the inclined plane is 4.

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Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at

netineya [11]

Answer:

Δu=1300kJ/kg

Explanation:

Energy at the initial state

$p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)$

Is saturated vapor at initial pressure we have

$p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)$

Process 2-3 is a constant volume process

$p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)$

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

$=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg$

Δu=1300kJ/kg

3 0

3 years ago

Phÿśïčš<br><br>write the relation between celsius and fahrenheit scale?

Aleksandr-060686 [28]

Answer:

C/100 = (F-32) / 180

or, C/5 = (F-32)/9

Explanation:

relation between any two scales is given by:

(X- lower fixed point ) / (upper fixed point -lower fixed point)

where X is any temperature

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3 years ago

Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut

Harman [31]

Answer: a) 11.76 m/s b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

$V_{f}=V_{o}+at$ (1)

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (2)

Where:

$V_{f}$ Is the final velocity of the object

$V_{o}=0$ Is the initial velocity of the object (it was dropped)

$a=9.8 m/s^{2}$ is the acceleration due gravity

$d$ is the height of the tower

$t=0.02min=1.2 s$ is the time it takes to the object to reach the ground

b) Begining with (1):

$V_{f}=0+at$ (3)

$V_{f}=at=(9.8 m/s^{2})(1.2 s)$ (4)

$V_{f}=11.76 m/s$ (5) This is the final velocity of the object

a) Substituting (5) in (2):

$(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d$ (6)

Clearing $d$ :

$d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})}$ (7)

$d=7.056 m$ (8) This is the height of the tower

4 0

4 years ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

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3 years ago

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Rom4ik [11]

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