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Mice21 [21]
3 years ago
5

The length of a simple pendulum is 0.65 m and the mass of the particle (the “bob”) at the end of the cable is 0.22 kg. The pendu

lum is pulled away from its equilibrium position by an angle of 7.6° and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What
is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob’s speed as it passes through the lowest point of the swing?
Physics
1 answer:
FromTheMoon [43]3 years ago
8 0

Answer:

A student is conducting a pendulum experiment. Which of the following pieces of safety equipment would be the most vital to conduct this test?eibisbbbhbcjb

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Alex777 [14]

Answer:

Chemical energy is converted to thermal energy and electrical energy.

Explanation:

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3 years ago
on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker
Aleksandr-060686 [28]
B4 the tackle: 

<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>

<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>


<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>

<span>The vector triangle is right angled: </span>

<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>

<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>

<span>v(f) = 5.6 m/s (to 2 sig figs) </span>


<span>direction of v(f) is the same as the direction of the final momentum </span>

<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>


<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>




<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
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3 years ago
What do we mean by the observable universe?
mote1985 [20]
The observable universe<span> is a spherical region of the </span>Universe, <span>comprising all matter that can be observed from Earth at the present time, because light and other signals from these objects have had time to reach Earth since the beginning of the cosmological expansion.


</span>
4 0
3 years ago
A car has uniformly accelerated from rest to a speed of 25m/s after traveling 75m. What is its acceleration in m/s^2
Roman55 [17]
<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’): </em></h2><h2><em> </em></h2><h2><em>The car’s speed is 25m/s </em></h2><h2><em>The distance travelled is 75m </em></h2><h2><em>Then we have the formulas for speed and distance: </em></h2><h2><em> </em></h2><h2><em>v = a x t -> 25 = a x t </em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2 </em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say: </em></h2><h2><em> </em></h2><h2><em>t = 25 / a </em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a </em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1): </em></h2><h2><em> </em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2 </em></h2><h2><em>t^2 = 150 / a </em></h2><h2><em>Since t has the same value for both truths we can say: </em></h2><h2><em> </em></h2><h2><em>625 / a^2 = 150 / a </em></h2><h2><em> </em></h2><h2><em>Thus multiply both sides with a^2: </em></h2><h2><em> </em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17 </em></h2><h2><em> </em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
4 0
3 years ago
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