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bija089 [108]
3 years ago
13

When a pendulum is swinging, the velocity is highest at which point?

Physics
2 answers:
strojnjashka [21]3 years ago
7 0
<h2>Answer:</h2>

<u>The velocity is highest </u><u>at the center, when the bob is hanging straight down </u>

<h2>Explanation:</h2>

As the pendulum swings to its highest position, its kinetic energy decreases to 0 on both right and left sides. If we tie a string to an object and let it swing back and forth, we will see that object’s velocity is 0 m/s at the highest points on the left and right sides and it is highest at at the center, when the bob is hanging straight down due to gravitational pull and the stored potential energy.

balu736 [363]3 years ago
4 0

Answer:

The velocity is highest at the lowest point of the trajectory

Explanation:

We can answer the question by applying the law of conservation of energy. In fact, neglecting friction and air resistance, the mechanical energy of the pendulum is constant during the motion:

E=K+U=const.

where:

K=\frac{1}{2}mv^2 is the kinetic energy, with m being the mass of the pendulum and v the velocity

U=mgh is the potential energy, with g being the acceleration due to gravity and h the heigth of the pendulum with respect to the lowest position

Since E must remain constant, we see that when K is maximum, U is minimum, and viceversa. This also means that when the velocity (v) is maximum, then the height (h) is minimum, so the pendulum must be at its lowest point.

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Momentum usually has the symbol p It is a vector, What is the correct way to
Nadya [2.5K]

Answer:

Add an arrow above the symbol p to show it is a vector. Sometimes it is italicized in textbooks.

Explanation:

6 0
3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}&#10; 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
A 1200 kg car accelerates from 13 m/s to 17 m/s. find the change in momentum of the car.
mixas84 [53]
P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
Δv= 17m/s-13m/s=4m/s
Now
p=mΔv
p=(1200kg)*(4m/s)
p=4800kgm/s


8 0
2 years ago
Please help me out! Thanks! (:
ElenaW [278]

X=Ft is a possible equation for X because by using SI units its final term became Kg metre per second

3 0
3 years ago
Time running out plzz hurry!!!!!!
Vesnalui [34]
I think you would hear a lower pitch
3 0
3 years ago
Read 2 more answers
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