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insens350 [35]
3 years ago
15

What are some forms of mechanical energy

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
4 0
Objects have mechanical energy<span> if they are in motion and/or if they are at </span>some<span> position relative to a zero potential </span>energy<span> position (for example, a brick held at a vertical position above the ground or zero height position). A moving car possesses </span>mechanical energy<span> due to its motion (kinetic </span>energy<span>).</span>
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In the following figure, if AB ǁ CD, then find the measure of PCD and CPD.​
mina [271]

Answer:

CPD = 80

PCD = 44

Explanation:

Given

AB || CD

BAD = 56

CPA = 100

See attachment

Required

Determine PCD and CPD

First, we need to calculate CPD

Since DPA is a straight line and CPA = 100;

We have that:

CPA + CPD = 180 --- angle on a straight theorem

Substitute 100 for CPA

100 + CPD = 180

Subtract 100 from both sides

100-100 + CPD = 180-100

CPD = 80

Next, we calculate PCD

We have that:

DAB= ADC = 56  --alternate angle

In triangle PCD

PCD + CPD + PDC = 180 --- angles in a triangle

Where

PDC = ADC = 56

So, we have:

PCD +80 + 56 = 180

PCD +136 = 180

Subtract 136 from both sides

PCD = 180 - 136

PCD = 44

6 0
3 years ago
It is a general exercise that we do before practicing sports activities<br><br> worm-up or cool down
Montano1993 [528]

It is called a warm-up.

3 0
3 years ago
Read 2 more answers
What is the mechanical advantage of a 8 m ramp that rises 2 m to a stage?
Neko [114]

Answer:

Mechanical advantage = 4

Explanation:

Given the following data;

Distance of effort, de = 8m

Distance of ramp, dr = 2m

To find the mechanical advantage;

Mechanical advantage = de/dr

Substituting into the equation, we have;

Mechanical advantage = 8/2

Mechanical advantage = 4

5 0
2 years ago
Due to historical difficulty in delivering supplies by plane, one of your colleagues has suggested you develop a catapult for sl
ikadub [295]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • We can take the initial velocity vector, which magnitude is a given (67 m/s) and project it along two directions perpendicular each other, which we choose horizontal (coincident with x-axis, positive to the right), and vertical (coincident with y-axis, positive upward).
  • Both movements are independent each other, due to they are perpendicular.
  • In the horizontal direction, assuming no other forces acting, once launched, the supply must keep the speed constant.
  • Applying the definition of cosine of an angle, we can find the horizontal component of the initial velocity vector, as follows:

       v_{avgx} = v_{o}*cos 50 = 67 m/s * cos 50 = 43.1 m/s (1)

  • Applying the definition of average velocity, since we know the horizontal distance to the target, we can find the time needed to travel this distance, as follows:

       t = \frac{\Delta x}{v_{avgx} } = \frac{400m}{43.1m/s} = 9.3 s  (2)

  • In the vertical direction, once launched, the only influence on the supply is due to gravity, that accelerates it with a downward acceleration that we call g, which magnitude is 9.8 m/s2.
  • Since g is constant (close to the Earth's surface), we can use the following kinematic equation in order to find the vertical displacement at the same time t that we found above, as follows:

       \Delta y = v_{oy}  * t - \frac{1}{2} *g*t^{2} (3)

  • In this case, v₀y, is just the vertical component of the initial velocity, that we can find applying the definition of the sine of an angle, as follows:

       v_{oy} = v_{o}*sin 50 = 67 m/s * sin 50 = 51.3 m/s (4)

  • Replacing in (3) the values of t, g, and v₀y, we can find the vertical displacement at the time t, as follows:

       \Delta y = (53.1m/s * 9.3s) - \frac{1}{2} *9.8m/s2*(9.3s)^{2} = 53.5 m (5)

  • Since when the payload have traveled itself 400 m, it will be at a height of 53.5 m (higher than the target) we can conclude that the payload will be delivered safely to the drop site.
4 0
3 years ago
A transverse wave is traveling from north to south. Which statement could be true for the motion of the wave particles in the me
Vikki [24]
Transverse waves travel on a direction that is perpendicular to the motion of the particles (or whatever medium is waving)  So the particles must be moving east to west, which is transverse to the north-south motion of the wave
3 0
3 years ago
Read 2 more answers
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