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Vikentia [17]
3 years ago
10

Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4

.50 cm in diameter? (b) At what speed will it shoot out if the di- ameter of the hole is three times as large?
Physics
1 answer:
kati45 [8]3 years ago
5 0

Answer:

velocity  = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = \frac{\pi }{4} × (4.50×10^{-2})²  × velocity

solve it we get

velocity  = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = \frac{\pi }{4} × 3 ×  (4.50×10^{-2})²  × velocity

velocity = 52.4 m/s

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3 years ago
A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

5 0
3 years ago
A 1,200 kg car travels at 20 m/s. what is it’s momentum ?
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3 years ago
Which statement best describes the relationship between atoms and nuclear forces? (2 points)
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4 0
2 years ago
Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an
ddd [48]

Answer:

Al's mass is 102.92  kg  

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

F_{net} = 0

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

F_{net} = \frac{dp_{horizontal}}{dt} = 0

p_{horizontal_i }= p_{horizontal_f}

where the suffix i and f means initial and final respectively.

The initial momentum will be:

p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}

But, as they are at rest, initially

p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0

p_{horizontal_i} = 0

So, this means:

p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0

We know that the have an combined mass of 195 kg:

m_{total} = m_{Al} + m_{Jo} = 195 \ kg.

so:

m_{Jo} = 195 \ kg - m_{Al}.

m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0

m_{Al} \  v_{Al_f} - m_{Al} \  v_{Jo_f}= - 195 \ kg \  v_{Jo_f}

m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}

m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } {  v_{Al_f} - v_{Jo_f} }

m_{Al} = \frac{195 \ kg  \ v_{Jo_f} } {    v_{Jo_f} - v_{Al_f} }

Now, we can use the values:

v_{Al_f}= 10.2 \frac{m}{s}

v_{Jo_f}= - 11.4 \frac{m}{s}

where the minus sign appears as they are moving at opposite directions

m_{Al} = \frac{195 \ kg  ( - 11.4 \frac{m}{s} ) } {   (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }

m_{Al} = 102.92 \ kg

and this is the Al's mass.

5 0
2 years ago
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