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umka21 [38]
3 years ago
11

Two identical gliders on an air track are held together by a piece of string, compressing a spring between the gliders. While th

ey are moving to the right at a common speed of 0.325 m/s, someone holds a match under the string and burns it, letting the spring force the gliders apart. One glider is then observed to be moving to the right at 1.33 m/s. What velocity does the other glider have
Physics
1 answer:
baherus [9]3 years ago
3 0

Answer:

Explanation:

Both the gliders have same mass and same speed of .325 m /s .

The detachment of spring will create equal and opposite force on both of them , one force adding momentum to one and the other force reducing the momentum of the other . Since mass of both the gliders are same , change in velocity too will be same . Let change in velocity be v .

for forward moving glider

.325 + v = 1.33

v = 1.005 m /s

The other glider will have final velocity

= .325 - 1.005

= - .68 m /s

So other glider will go in opposite direction with velocity of .68 m /s .

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Appeal system enables the higher court to review the case once again. Possible situation for appeal arises when the defendant consider punishment very tough.

<u>Explanation:</u>

The Appeal system is very important in the legal system. Appeal means a process which helps in reviewing the cases. In case of the appeal, a request is made to a higher court to look into the case once again.

This process helps those who feel that the earlier judgment in case of the case was not correct.

The Possible situation where someone needs to appeal is when the defendant is found guilty and we feel that the punishment given is very harsh. This is possible mostly in criminal cases.

6 0
3 years ago
Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show
Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

m_2 = \frac{20}{3} kg

Part b)

Angular acceleration is given as

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = 176.6 N

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

m_1g r_1 = m_2 g r_2

20 g(0.5) = m_2 g(1.5)

10 = 1.5 m_2

m_2 = \frac{20}{3} kg

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

m_1g - T = m_1 a

also we have

T r_1 = I\alpha

now we have

m_1g = \frac{I a}{r_1^2} + m_1 a

a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}

a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}

a = 0.981 m/s^2

so angular acceleration is given as

\alpha = \frac{a}{r_1}

\alpha = \frac{0.981}{0.5}

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = \frac{I\alpha}{r_1}

T = \frac{45 (1.96)}{0.5}

T = 176.6 N

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3 years ago
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Gnoma [55]
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3 years ago
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Answer:

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3 years ago
Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum
marta [7]

Answer:

  • v_{top} = 61.96 \frac{mi}{h}

Explanation:

To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

We know that the top speed is

v_{top} = 27.7 \frac{m}{s}

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

1609.34 \ m = 1 \ mi

Now, we can divide by 1609.34 meters on both sides:

\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}

The left sides equals 1, so

1 = \frac{1 \ mi}{ 1609.34 \ m}

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}

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This is the top speed in miles per second, now, for obtaining miles per hour:

<h3>hours from seconds</h3>

We can do pretty much the same, first, the equivalence:

1 \ h = 3600 \ s

as the seconds are dividing in the velocity, we know divide by 1 hour.

\frac{1 \ h}{ 1 \ h} = \frac{3600 \ s}{ 1 \ h}

1 = \frac{3600 \ s}{ 1 \ h}

and know we just multiply our top speed by this conversion factor

v_{top} = 0.0172120 \frac{mi}{s}  \frac{3600 \ s}{ 1 \ h}

v_{top} = 61.96 \frac{mi}{h}

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3 years ago
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