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VLD [36.1K]
2 years ago
6

Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving

at speed vi on a frictionless surface while attached to an ideal, horizontally fixed spring with a force constant of ki . Let x denote the displacement of the block from its equilibrium position. Order the systems from largest total mechanical energy to smallest.
a. m1= 0.5KG k2=500 N/m amplitude A = 0.02 m
b. m2= 0.6KG k2=300 N/m v2= 1 m/s . when passing through equilibrium
c. m3= 1.2KG k3=400 N/m v3= 0.5 m/s . when passing through x= -0.01 m
d. m4= 2 KG k4=200 N/m v4= 0.2 m/s . when passing through x=-0.05 m
Physics
1 answer:
Rzqust [24]2 years ago
4 0

Answer:

The order is 2>4>3>1 (TE)

Explanation:

Look up attached file

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You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba
Cloud [144]

Answer:

The ball reaches Barney  head in  t = 8 \ s

Explanation:

From the question we are told that

 The rise velocity is  v  =  14.70 \  m/s

  The height considered is h =  196 \  m

   The horizontal velocity of the large object is  v_h  =  8.50 \  m/s

   

Generally from kinematic equation  

   s = ut + \frac{1}{2} gt^2

Here s is the distance of the object from Barney head ,

        u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter

So  

     u = -14.7 m/s

So

    196  = -14.7 t  + \frac{1}{2} * 9.8 * t^2

=  4.9 t^2 - 14.7t - 196 = 0

Solving the above equation using quadratic formula  

    The value of  t obtained is  t = 8 \ s

6 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th
kolbaska11 [484]

Answer:

mg=200.4 N.

Explanation:

This problem can be solved using Newton's law of universal gravitation: F=G\frac{m_{1}m_{2}}{r^{2}},

where F is the gravitational force between two masses m_{1} and m_{2}, r is the distance between the masses (their center of mass), and G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2}) is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting w_{s} be the weight on the surface:

w_{s}=mg,

mg=8*10^{2},

m=(8*10^{2})/g,

since we now that g=9.8m/s^{2} we get that the mass is

m=81.6kg.

Now we can use Newton's law of universal gravitation

F=G\frac{Mm}{r^{2}},  

where m is the mass of the astronaut and M is the mass of the earth. From Newton's second law we know that

F=ma,

in this case the acceleration is the gravity so

F=mg, (<u>becarefull, gravity at this point is no longer</u> 9.8m/s^{2} <u>because we are not in the surface anymore</u>)

and this get us to

mg=G\frac{Mm}{r^{2}}, where mg is his new weight.

We need to remember that the mass of the earth is M=5.972*10^{24}kg and its radius is 6.37*10^{6}m.

The total distance between the astronaut and the earth is

r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6} meters.

Now we can compute his weigh:

mg=G\frac{Mm}{r^{2}},

mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}},

mg=200.4 N.

5 0
2 years ago
Determine the change in thermal energy of 100 g of copper (M = 63,5, Debye 348K) if it is cooled from
Setler [38]

Answer:

given,

mass of copper = 100 g

latent heat of liquid (He) = 2700 J/l

a) change in energy

Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (300 - 4)

Q = 11153.63 J

He required

Q = m L

11153.63 = m × 2700

m = 4.13 kg

b) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (78 - 4)

Q = 2788.41 J

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Q = m L

2788.41 = m × 2700

m = 1.033 kg

c) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (20 - 4)

Q = 602.90 J

He required

Q = m L

602.9 = m × 2700

m =0.23 kg

8 0
2 years ago
In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.
denis23 [38]
The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.
5 0
2 years ago
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