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Andre45 [30]
3 years ago
5

Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t

he point of ejection are 1 meter/second and 0.5 meters/second respectively, calculate the ratio of their horizontal distances of impact. 1:04 1:234 2:01 1.41:1 3:02
Physics
2 answers:
givi [52]3 years ago
8 0
For t1:

t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec

For t2:

t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec

Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)

d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m

Where:

d = hor. distance

ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25

The answer is 3.19 : 2.25
LuckyWell [14K]3 years ago
3 0

Answer:

1.41 : 1

Explanation:

time required to reach the ground is given as

t = \sqrt{\frac{2h}{g}}

now we know that

h = 50 cm and h = 100 cm

so the ratio of time of two is given as

\frac{t_1}{t_2} = \sqrt{\frac{50}{100}}

\frac{t_1}{t_2} = \sqrt{\frac{1}{2}}

now the distance traveled by the object is given as

d = vt

now the ratio of the distance moved is given as

\frac{d_1}{d_2} = \frac{v_1t_1}{v_2t_2}

\frac{d_1}{d_2} = \frac{1}{0.5}\times \sqrt{\frac{1}{2}}

\frac{d_1}{d_2} = \sqrt 2 = 1.41 : 1

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Answer:

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Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

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\omega =\frac{v}{R}

The kinetic energy of the system in terms of individual mass can be written as:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2

By replacing \omega with \frac{v}{R} ; we have:

K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2

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Equating both equation (1) and (2); we have:

m_e = m_1+m_2+\frac{I}{R^2}

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The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by  the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

f_{block }=m_1gsin \beta

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

f_{cylinder} = m_2gsin \phi

Equating the above both equations; we have the equation of motion of the  equivalent system to be

m_e \bar x = f_{cylinder}-f_{block}

which can be written as follows from the previous derivations

(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

Finally; the equation of the motion of the block of mass m_1 in terms of its displacement is = (m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)

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