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3 years ago

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A crate with a mass of 193.5 kg is suspended from the end of a uniform boom with a mass of 91.7 kg. The upper end of the boom is

supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall. Calculate the tension in the cable.

1 answer:

tiny-mole [99]3 years ago

4 0

Answer:

2538.72 N

Explanation:

Given data:

There are some missing values, so, I will assume them, and then you affix them. It's basically just substitution....

m₁ = mass of crate = 193.5 kg

m₂ = mass of the boom = 91.7 kg

g = acceleration by gravity = 9.81 m/s²

L = length of the boom

θ = angle of the boom to the horizontal

φ = angle of the cable to horizontal

Solution:

I'm assuming that

tan (θ) = 6/12 = 0.5

This then makes

θ = arctan(0.5)= 26.57°

The second and last assumption I'm making is that

tan(φ) = 3/12 = 0.25

This also makes

φ = arctan(0.25) = 14.04°

Now, since the system is in equilibrium, the magnitude of clockwise torque has to be equal to anticlockwise torque. This means that

m₁ * g * cos(θ) *L/2 + m₂ * g *cos(θ) * L = T * sin(θ+φ) * L

(m₁/2 + m₂) * g * cos(θ) = T * sin(θ+φ), making T subject of formula, we have

T = (m₁/2 + m₂) * g * cos(θ) / sin(θ + φ)

T = [(193.5)/2 + (91.7)] * (9.81) * cos(26.57°) / sin(26.7°) + (14.04°))

T = [(96.75 + 91.7) * 9.81 * 0.894] / 0.651

T = 188.45 * 8.77 / 0.651

T = 1652.7065 / 0.651

T = 2538.72 N

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Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth’s center. Satellite A is to o

Vera_Pavlovna [14]

Answer:

Explanation:

Orbital radius of satellite A , Ra = 6370 + 6370 = 12740 km

Orbital radius of satellite B , Rb = 6370 + 19110 = 25480 km

Orbital potential energy of a satellite = - GMm / r where G is gravitational constant , M is mass of the earth and m is mass of the satellite

Orbital potential energy of a satellite A = - GMm / Ra

Orbital potential energy of a satellite B = - GMm / Rb

PE of satellite B /PE of satellite A

= Ra / Rb

= 12740 / 25480

= 1 / 2

b ) Kinetic energy of a satellite is half the potential energy with positive value , so ratio of their kinetic energy will also be same

KE of satellite B /KE of satellite A

= 1 / 2

c ) Total energy will be as follows

Total energy = - PE + KE

- P E + PE/2

= - PE /2

Total energy of satellite B / Total energy of A

= 1 / 2

Satellite B will have greater total energy because its negative value is less.

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3 years ago

Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through polystyrene. The waveleng

klemol [59]

Answer:

Speed:

$2.01x10^{8}m/s$

Wavelength:

$4.24x10^{-7}m$

Frequency:

$4.74x10^{14}Hz$

Explanation:

The speed of the laser as it travels through polystyrene can be determine by means of the equation of the refraction index:

$n = \frac{c}{v}$ (1)

Where c is the speed of light and v is the speed of the laser in the medium.

Therefore, v will be isolated from equation 1

$v = \frac{c}{n}$

$v = \frac{3x10^{8}m/s}{1.490}$

$v = 2.01x10^{8}m/s$

Hence, the speed of the laser has a value of $2.01x10^{8}m/s$

Frenquency:

Since, wavelength is the only one who depends on the media. Therefore the frequency in both medium will be the same.

To determine the frequency it can be used the following equation

$c = \nu \cdot \lambda$ (2)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength

Then, $\nu$ wil be isolated from equation 2.

$\nu = \frac{c}{\lambda}$ (3)

Before using equation 3 it is necessary to express $\lamba$ in units of meters.

$\lambda = 632.8nm . \frac{1m}{1x10^{9}nm}$ ⇒ $6.328x10^{-7}m$

$\nu = \frac{3x10^{8}m/s}{6.328x10^{-7}m}$

$\nu = 4.74x10^{14}s^{-1}$

$\nu = 4.74x10^{14}Hz$

Hence, the frequency of the laser has a value of $4.74x10^{14}Hz$

Wavelength:

To determine the wavelength it can be used:

$v = \nu \cdot \lambda$

$\lambda = \frac{v}{\nu}$

Where v is the speed of the laser through the polystyrene.

$\lambda = \frac{2.01x10^{8}m/s}{4.74x10^{14}s^{-1}}$

$\lambda = 4.24x10^{-7}m$

Hence, the wavelength of the laser has a value of $4.24x10^{-7}m$

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3 years ago

Identify the independent, dependent, and constant variables for different experiments.

diamong [38]

ANSWER:

IV, Type of dish detergent. DV, height of foam. CV, type of container, amount of water in container, temperature of water, time the container is agitated.

Explanation:

Independent variable(IV)- what you change during the experiment.

dependent variable(DV)- what you're measuring during an experiment. The dependent variable is DEPENDENT because it's results DEPEND on the independent variable at play.

Constant variables(CV)- things that do not change in order to isolate the tested variables as much as possible.

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3 years ago

The acceleration due to gravity on Jupiter is 2.5 times what it is here on earth. An object weighing 347.9 N here on earth will

inessss [21]

Answer: weight on Jupiter = 869.75 N

mass on Earth = mass on Jupiter = 35.5 Kg

Explanation:

W = mg

W = weight

m = mass

g = gravitational acceleration [ on the Earth, g₁ = 9,8 N/kg ]

On the Earth,

G₁ = m x g₁ = 347,9 N

On the Jupiter,

G₂ = mg₂

mass on the Earth = mass on the Jupiter !

m = G₁ : g = 347.9 N : 9,8 N/kg = 35.5 kg

G2 : G1 = 2.5

G₂ = 2,5 G₁ = 2,5 x 347.9 N = 869,75 N

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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