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anastassius [24]

4 years ago

11

The small piston of a hydraulic lift has a crosssectional area of 2.23 cm2 and the large piston 297 cm2 . What force must be app

lied to the small piston for the lift to raise a load of 2.4 kN? (In service stations, this force is usually exerted by compressed air.) Answer in units of N.

1 answer:

Lerok [7]4 years ago

8 0

Answer:

The force that must be applied to the small piston = 18.02 N.

Explanation:

Hydraulic Press: This is a device that produce a very large force to compress something e.g printing press.

From pascal's principle,

In an hydraulic press,

f/a = F/A......................... Equation 1

Where f = force applied to the small piston, a = area of the small piston, F = force required to lift the load/ force produced at the large piston. A = Area of the large piston.

Making f the subject of the equation above,

f = F×a/A.............................. Equation 2

Given: F = 2.4 kN = 2400 N, a = 2.23 cm², A = 297 cm².

Substituting into equation 2

f = 2400(2.23)/297

f = 18.02 N.

Thus the force that must be applied to the small piston = 18.02 N.

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An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down

soldier1979 [14.2K]

Answer:

$1.9\times 10^{-4}$

$1.2\times 10^{-4}\ m$

Explanation:

r = Radius = 2.7 cm

F = Force = $3.2\times 10^4\ N$

A = Area = $\pi r^2$

$\sigma$ = Stress = $\frac{F}{A}$

E = Young's modulus = $7\times 10^{10}\ Pa$

$\epsilon$ = Strain

$L_0$ = Original length = 67 cm

$\Delta L$ = Change in length

Young's modulus is given by

$E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}$

Strain is $1.9\times 10^{-4}$

Strain is given by

$\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m$

The cylinder height decreases by $1.2\times 10^{-4}\ m$

3 0

3 years ago

A car accelerates from zero to a speed of 110

Verizon [17]

The car's rate of acceleration : a = 2.04 m/s²

<h3>Further explanation</h3>

Given

speed = 110 km/hr

time = 15 s

Required

The acceleration

Solution

110 km/hr⇒30.56 m/s

Acceleration is the change in velocity over time

a = Δv : Δt

Input the value :

a = 30.56 m/s : 15 s

a = 2.04 m/s²

3 0

3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span> $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
<span>where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span> $E=K_{max} = \frac{1}{2}mv_{max}^2$
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span> $E=U_{max}= \frac{1}{2}k A^2$
<span>
Since the mechanical energy must be conserved, we can write
</span> $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
<span>from which we find the maximum speed
</span> $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
</span> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span> $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
<span>And since
</span> $E=U+K$
<span>we find the kinetic energy when the glider is at this position:
</span> $K=E-U=0.36 J - 0.05 J = 0.31 J$
<span>And then we can find the corresponding velocity:
</span> $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span> $a(t)=\omega^2 x(t)$
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
</span> $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

4 years ago

How do two sublevels of the same principal energy level differ from each other

Colt1911 [192]

Each energy sublevel corresponds to an orbital of a different shape.

Explanation:

Two sublevels of the same principal energy level differs from each other if the sublevels corrresponds to an orbital of a different shape.

The principal quantum number of an atom represents the main energy level in which the orbital is located or the distance of an orbital from the nucleus. It takes values of n = 1,2,3,4 et.c
The secondary quantum number gives the shape of the orbitals in subshells accommodating electrons.
The number of possible shapes is limited by the principal quantum numbers.

Take for example, Carbon:

1s² 2s² 2p²

The second energy level is 2 but with two different sublevels of s and p. They have different shapes. S is spherical and P is dumb-bell shaped .

Learn more:

Quantum number brainly.com/question/9288609

#learnwithBrainly

7 0

4 years ago

I don’t know this one I need help

nikdorinn [45]

D
Using the kinetic energy 1/2mv^2 formula
5*10^5 is the answer

4 0

2 years ago