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trapecia [35]
3 years ago
13

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.2 m. The mug s

lides off the counter and strikes the floor 1.35 m from the base of the counter. Assume the mug moves in the positive x-direction. Give your answer in three significant figures.
a) With what velocity did the mug leave the counter?
b) What was the direction (in degrees and with respect to the horizontal) of the mug's velocity just before it hit the floor?
Physics
1 answer:
ZanzabumX [31]3 years ago
6 0

Part a can be solve using the equation of trajectory:

Y = x tana + (g*x^2)/ [2(V0^2)*(cos a)^2]

Where y is the height

X is the length

G is the acceleration due to gravity

Vo Is the initail velocity

a is the angle of trajectory

1.2 = 1.35  tan(0) + (9.81*1.35^2)/ [2(V0^2)*(cos 0)^2]

Solve for V0 = 2.729 m/s

b. can be solve using the formula

v = sqrt(2gy)

= sqrt ( 2*1.2*9.81)

= 4.852 m/s  going down ( 0 degree from the horizontal)

<span> </span>

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3 years ago
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The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr
Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V=A_1V_1+ A_2V_2+A_3V_3

where suffix 1,2,3 refers to the three pipes.

            =27L/min+16L/min+11 L/min

            =54L/min

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   d= 2 cm

\frac{\pi }{4} d^2 x v = 54

v= \frac{4}{\pi } x \frac{54}{2^2}

-> A_1 x V_1=27L/min => \frac{\pi }{4} d_1^2 x v_1

d_1= 1.3cm

\frac{\pi }{4} d^2 x v_1 = 27

v_1= \frac{4}{\pi } x \frac{27}{1.3^2}

Next is to find the ratio of speed i.e \frac{v}{v_1}

\frac{4}{\pi } x \frac{54}{2^2} / \frac{4}{\pi } x \frac{27}{1.3^2} => \frac{54}{27} \frac{1.3^2}{2^2}

\frac{v}{v_1}= 0.845

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Water that flows across the surface is called a;

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Answer:

288N

Explanation:

Given parameters:

Mass of Cheetah = 12kg

Acceleration  = 24m/s²

Unknown:

Force needed by the cheetah to run  = ?

Solution:

The force needed by the Cheetah to run is the net force.

According to Newton's law;

    Force  = mass x acceleration

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