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Nezavi [6.7K]
3 years ago
10

What can you conclude about the total mechanical energy of a pendulum as it swings back and forth?

Physics
1 answer:
Alchen [17]3 years ago
8 0

Answer:

The total mechanical energy of a pendulum is conserved neglecting the friction.

Explanation:

  • When a simple pendulum swings back and forth, it has some energy associated with its motion.
  • The total energy of a simple pendulum in harmonic motion at any instant of time is equal to the sum of the potential and kinetic energy.
  • The potential energy of the simple pendulum is given by P.E = mgh
  • The kinetic energy of the simple pendulum is given by, K.E = 1/2mv²
  • When the pendulum swings to one end, its velocity equals zero temporarily where the potential energy becomes maximum.
  • When the pendulum reaches the vertical line, its velocity and kinetic energy become maximum.
  • Hence, the total mechanical energy of a pendulum as it swings back and forth is conserved neglecting the resistance.
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A 34.0 %-efficient electric power plant produces 800 MW of electric power and discharges waste heat into 20∘C ocean water. Suppo
mojhsa [17]

Answer:

77647

Explanation:

\eta = Efficiency = 34%

Power used in 1 home = 0.02 MW

Total power is

P=\dfrac{800}{\eta}\\\Rightarrow P=\dfrac{800}{0.34}\\\Rightarrow P=2352.94117\ MW

Waste of power

2352.94117-800=1552.94117\ W

Number of homes would be given by

n=\dfrac{1552.94117}{0.02}=77647.0585\ homes

The number of homes that could be heated with the waste heat of this one power plant is 77647

8 0
2 years ago
a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleratio
Harman [31]

Answer:

The  angular acceleration is  \alpha  = 3.235 \ rad/s ^2

Explanation:

From the question we are told that

    The moment of inertia is  I  =  0.034\ kg \cdot m^2

     The  net torque is  \tau  =  0.11\ N \cdot m

Generally the net torque is mathematically represented as

           \tau =  I  *  \alpha

Where \alpha is the angular acceleration so  

        \alpha  =  \frac{\tau }{I}

substituting values

         \alpha  =  \frac{0.1 1}{ 0.034}

        \alpha  = 3.235 \ rad/s ^2

6 0
3 years ago
Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec
sammy [17]

Answer:

1) The maximum jump height is reached at A. 0.337s

2) The maximum center of mass height off of the ground is B. 1.64m

3) The time of flight is C. 0.834s

4) The distance of jump is B. 7.49m

Explanation:

First of all we need to decompose velocity in its rectangular components, so

v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s

1) We use, v_{fy}=v_{iy}-gt, as we clear it for t and using the fact that v_{fy}=0 at max height, we obtain t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s

2) We can use the formula y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2} for t=0.337s, so

y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m

3) We can use the formula y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find total time of fligth, so 0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66, as it is a second-grade polynomial, we find that its positive root is t=0.834s

4) Finally, we use x=v_{x}t=8.05m/s(0.834s)=6.71m, as it has an additional displacement of 0.77m due the leg extension we obtain,

x=6.71m+0.77m=7.48m, aprox x=7.49m

5 0
3 years ago
A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft
elena-s [515]

Answer:

The water level rises more when the cube is located above the raft before submerging.

Explanation:

These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.

Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.

When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.

Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]

Density is given by:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyancy force can be calculated using the following equation:

F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]

Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.

7 0
3 years ago
The force applied when using a simple machine
jenyasd209 [6]

Answer:

effort

Explanation:

effort is the answer

7 0
3 years ago
Read 2 more answers
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