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White raven [17]
1 year ago
10

You redo the primitive yo yo experiment (Figure 1), but instead of holding the free end of the string stationary, you move your

hand vertically so that the tension in the string equals 2M /3. What is the vertical acceleration of the yo yo's cena of mass? Does it accelerate upward or downward? Express your answer to three significant figures and include the appropriate units. Enter positive value of the acceleration is directed upward and negative value if the acceleration is directed downward.
Physics
1 answer:
zhannawk [14.2K]1 year ago
5 0

The vertical acceleration of the yo yo cent  of mass is 0.66 m/s².

We know that, T = M*a

And T is given as 2M/3

2M/3 = M*a

So, a = 0.66 m/s²

What is Acceleration ?

Acceleration is the general term for any process where the velocity changes. There are only two ways to accelerate either by increasing the speed or decreasing direction, or both. The reason for this is that velocity includes both a speed and a direction.

You cannot possibly be accelerating if you don't also change you direction and speed, regardless of how swiftly you are travelling. Due to this, a jet experiences no acceleration even when it is moving at a high speed. In this case, 800 miles per hour, because its velocity is constant.

When it lands, the jet will accelerate as it slows down and quickly come to a stop.

Learn more about Acceleration from given link

brainly.com/question/605631

#SPJ4

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The image seems to be behind the mirror, but nothing is really there.

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What are the two parts of a force pair?
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8 0
2 years ago
An airplane is moving at 350 km/hr. If a bomb is
bogdanovich [222]

Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
A Ferris wheel has diameter of 10 m. It rotates at a uniform rate and makes one revolution in 8.0 seconds. A person weighing 670
Nikolay [14]

Answer:  459.14 N

Explanation:

from the question, we have

diameter = 10 m

radius (r)  = 5 m

weight (Fw) = 670 N

time (t) = 8 seconds

Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path, therefore we apply the equation below

∑ F = F c =  F w − Fn ..............equation 1

Fn = Fw − Fc = mg − (mv^2 / r) ...................equation 2

substituting the value of v as (2πr / T) we now have

Fn = mg − (m(2πr / T )^2) / r

Fn= mg − (4(π^2)mr / T^2)   ..........equation 3

Fw (mass of the person) = mg

therefore m = Fw / g

                m = 670 / 9.8 = 68.367 kg

now substituting  our values into equation 3

Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)

Fn = 670 - 210.86

Fn = 459.14 N

4 0
3 years ago
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