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soldi70 [24.7K]
3 years ago
8

Jacob is traveling at 5.00 m/s North. Jacob throws a ball with a velocity of 5.00 m/s South. Jacob throws the ball from a height

of 1.45 m above the ground. Assuming gravitational acceleration is 9.81 m/s^2, how long does the ball stay in the air after it is thrown? Describe the motion of the ball relative to the ground. Describe the motion of the ball relative to Jacob.
Physics
1 answer:
Nataly [62]3 years ago
3 0

Answer:t=0.54 s

Explanation:

Given

Jacob is traveling 5 m/s in North direction

Jacob throw a ball with a in south direction with a velocity of 5 m/s

Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0

Time taken by ball to reach ground

s=ut+\frac{gt^2}{2}

1.45=0+\frac{9.81\times t^2}{2}

t^2=frac{2\times 1.45}{9.81}

t=0.54 s

Motion of ball will be straight line

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Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 8=50.24rad/sec

Velocity of the string wave is equal to v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec

Power of wave propagation is equal to P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt

So power of the wave will be equal to 5.464 watt

6 0
2 years ago
a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame
natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}

where;

L = length of the bed

\mu = viscosity

U = superficial velocity

\epsilon = void fraction

dp = equivalent spherical diameter of bed material (m)

\rho = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6       --------------   equation (1)

24A + 576B = 24.1    ---------------  equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

       B = 0.017014

From equation (1)

12A + 144B  = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

A = \frac{9.6 -144(0.017014}{12}

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0
3 years ago
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Answer:

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7 0
2 years ago
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Identify which best describes the energy used to pluck guitar strings to make sound.
Novay_Z [31]
<span>The correct answer is: Mechanical Energy

Explanation:
As the guitar strings are plunked, the potential energy stored in the strings has an ability to make them vibrate. When the strings are vibrating, that potential energy is actually converted to the kinetic energy. Hence, the whole phenomena contains both the kinetic energy and the potential energy. The sum of kinetic energy and the potential energy is called Mechanical energy. Therefore, the correct answer is Mechanical Energy.</span>
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3 years ago
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4. Going back to the dog whistle in question 1, what is the minimum riding speed needed to be able to hear the whistle? Remember
jeyben [28]

Answer:

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7  m/s

Explanation:

The Doppler shift equation is given as follows;

f' = \dfrac{v - v_o}{v + v_s} \times f

Where:

f' = Required observed frequency = 20.0 kHz

f = Real frequency = 21.0 kHz

v = Sound wave velocity = 330 m/s

v_o = Observer velocity = X m/s

v_s = Source velocity = 0 m/s (Assuming the source is stationary)

Which gives;

20 = \dfrac{330- v_o}{330+0} \times 21

330 - v_o = (20/21)*330

v_o = 330 - (20/21)*330 = 15.7 m/s

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7  m/s.

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2 years ago
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