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insens350 [35]
3 years ago
6

The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses

Physics
1 answer:
Orlov [11]3 years ago
8 0
Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T =  tension over the frictionless pulley.

Write the equations of motion.
m₂g - T = m₂a            (1)
T - m₁g = m₁a            (2)

Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a

Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a

Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)

With  = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962

Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).

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An airplane starts from rest and accelerates at 10.8 m/s2 . what is its speed at the end of a 400 m long runway?
Cloud [144]

The final speed of an airplane is v = 92.95 m/s

The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.

Solution-

Here given,

Acceleration a= 10.8 m/s2 .

Displacement (s)= 400m

Then to find final speed of airplane v=?

Therefore from equation of motion can be written as,

v²=u²+ 2as

where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.

v²= 2×10.8 m/s²×400m

v²=8640m/s

v=92.95m/s

hence the final speed of airplane v =92.95m/s

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5 0
2 years ago
A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas
Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, \theta=33^{\circ}

Time for which the player's foot is in contact with it, \Delta t = 5.1\times 10^{-2}\ s

Part A,

The x component of the soccer ball's change in momentum is given by :

\Delta p_x=mv\ cos\theta

\Delta p_x=0.425\times 15\ cos(33)

p_x=5.34\ kg-m/s

The y component of the soccer ball's change in momentum is given by :

\Delta p_y=mv\ sin\theta

\Delta p_y=0.425\times 15\ sin(33)

p_y=3.47\ kg-m/s

Hence, this is the required solution.

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3 years ago
What is the resultant of a pair of forces, 100N, upward and 75N, downward?
soldi70 [24.7K]

Answer:

25N

Explanation:

100 - 75 = 25

That should be right if im not dumb...

3 0
3 years ago
A basketball has a mass of 567 g. Heading straight downward, in the direction, it hits the floor with a speed of 2 m/s and rebou
Vsevolod [243]

Answer:

\Delta p=2.27\frac{kg\cdot m}{s}

Explanation:

The momentum change is defined as:

\Delta p=p_f-p_i\\\Delta p=mv_f-mv_i\\\Delta p=m(v_f-v_i)(1)

Taking the downward motion as negative and the upward motion as positive, we have:

v_f=2\frac{m}{s}(2)\\v_i=-2\frac{m}{s}(3)

Replacing (2) and (3) in (1):

\Delta p=567*10^{-3}kg(2\frac{m}{s}-(-2\frac{m}{s}))\\\Delta p=2.27\frac{kg\cdot m}{s}

5 0
3 years ago
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