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Damm [24]
3 years ago
9

a dock worker pushes a 150.0 kg crate up a ramp that is 3.00 m high and 10.00 m long into the cargo bay of a ship. He exerts a 8

00.0 N force parallel to the ramp. find both the mechanical advantage and ideal mechanical advantage of the ramp. what is the efficiency of the ramp? ​
Physics
1 answer:
juin [17]3 years ago
3 0

Efficiency: (2940 J / 3250 J) x 100% = 90.46%

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A light source has a frequency of 7.56 x 1014 Hz. What is the wavelength, in nanometers (nm), of this light when it passes throu
antiseptic1488 [7]

Answer:

2nm

Explanation:

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The average kinetic energy of uf6(g) is 4.98 kj/mol at some temperature. what is the average kinetic energy of sf6(g) at that sa
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5 0
3 years ago
Read 2 more answers
(b) A cylinder of cross-sectional area 0.65m2 and
VLD [36.1K]

Answer:

The volume of the cavity is 0.013m^3

Explanation:

To find the volume of the cavity, the major parameter missing is the diameter of the cavity itself. we can obtain this using the following steps:

Step one:

Obtain the volume of the cylinder by dividing the mass of the cylinder by the density.

Volume of the cylinder = 2.1 / 11.053 =0.19m^{3}

Step two:

From the volume of the cylinder, we can get the radius of the cylinder.

radius = \sqrt{\frac{V}{\pi \times h}}  = \sqrt{\frac{0.19}{\pi \times 0.32}} =0.44m

Step three:

From the cross-sectional area, we can obtain the radius of the cavity.

Let the radius of the cavity be = r, while the radius of the cylinder be = R

CSA of cavity =

\pi({R^2}-r^2) = CSA\\0.65 = \pi (0.32^2-r^2)\\r= 0.115m

Step Four:

calculate the volume of the cavity using volume =\pi r^2 \times h

Recall that the cavity has the same height as the original cylinder

volume = \pi \times 0.115^2\times 0.32= 0.013m^3

8 0
3 years ago
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Brilliant_brown [7]

Answer:

Explanation:

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6 0
3 years ago
An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the
Alexxx [7]
(1) The position around equilibrium of an object in simple harmonic motion is described by
x(t) = A \cos (\omega t)
where
A is the amplitude of the motion
\omega is the angular frequency.

The velocity is the derivative of the position:
v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)
where 
v_0 = \omega A is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)
where
a_0=\omega^2 A is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
v_0 = \omega A = 4.3 m/s
a_0 = \omega^2 A = 0.65 m/s^2
From the first equation, we get
A= \frac{4.3 }{\omega} (1)
and if we substitute this into the second equation, we find the angular fequency
\omega=0.15 rad/s
while the amplitude is  (using (1)):
A=28.7 m


(b) We found in the previous step that the angular frequency of the motion is
\omega=0.15 rad/s
But the angular frequency is related to the period by
T= \frac{2 \pi}{\omega}
and so, the period is
T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s
5 0
3 years ago
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