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3 years ago

!!!!BRAINLIEST!!!!!

Physics

2 answers:

kykrilka [37]3 years ago

7 0

Answer:

The jp2003parker guy is extremely wrong

So he says that the size wont matter and a physical change should occur, but how would the size change without having a physical change occur.

Explanation:

GaryK [48]3 years ago

6 0

Answer:

no it doesnt

Explanation:

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Water from a vertical pipe emerges as a 10-cm-diameter cylinder and falls straight down 7.5 m into a bucket. The water exits the

cupoosta [38]

The diameter of the column of the water as it hits the bucket is 4.04 cm

The equation of continuity occurs in the fluid system and it asserts that the inflow and the outflow of the volume rate at the inlet and at the outlet of the system are equal.

By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;

Using the kinematics equation:

$\mathbf{v_f ^2 = v_i^2 + 2gh}$

$\mathbf{v_f ^2 =(2.0)^2 + 2\times 9.8 \times 7.5}$

$\mathbf{v_f ^2 =151 m/s}$

$\mathbf{v_f =\sqrt{151 m/s}}$

$\mathbf{v_f =12.29 \ m/s}$

From the equation of continuity:

$\mathbf{A_iV_i = A_fV_f}$

$\mathbf{\pi r^2_iV_i = \pi r^2_fV_f}$

$\mathbf{ r^2_iV_i = r^2_fV_f}$

$\mathbf{ (\dfrac{10}{2})^2\times 2.0 = r_f^2 \times 12.29}$

$\mathbf{ 50 = 12.29 \times r_f^2}$

$\mathbf{ r_f= \sqrt{\dfrac{50}{12.29} }}$

$\mathbf{ V_f= 2.02 \ cm }$

Since diameter = 2r;

∴

The diameter of the column of the water is:

= 2(2.02) cm

= 4.04 cm

Learn more about the equation of continuity here:

brainly.com/question/10822213

6 0

2 years ago

A bowling ball is 21.6 cm in diameter. What is the angular speed of these ball whenit is moving at 3.0 m/s?

sergejj [24]

Answer:

Angular speed = 27.78 rad/s (Approx)

Explanation:

Given:

Diameter = 21.6 cm

Speed = 3 m/s

Find:

Angular speed

Computation:

Radius = 21.6 / 2 = 10.8 cm = 0.108 m

Angular speed = v / r

Angular speed = 3 / 0.108

Angular speed = 27.78 rad/s (Approx)

3 0

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Vlad [161]

Answer: $7.95^{\circ}C$

Explanation:

Given

Mass of water is $m_1=1\ kg$

mass of ice is $m_2=0.1\ kg$

Latent heat of fusion $L=3.33\times 10^5\ kJ/kg$

The heat capacity of water is $c_{w}=4186\ J/kg^{\circ}C$

Suppose water is at $T^{\circ} C$ and it reaches to $0^{\circ}C$ to melt the ice

the heat released by water must be equivalent to heat absorbed by the ice

$\therefore \quad m_1c_w(T-0)=m_2\times L\\\Rightarrow 1\times 4186\times T=0.1\times 3.33\times 10^5\\\Rightarrow T=7.95^{\circ}C$

6 0

2 years ago

A 2.02 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.22 kΩ resistor. (a) Calculate the current in the

Mandarinka [93]

Answer:

a) 0.048A

b) 0.18µC

c) 1.85A

Explanation:

The discharged current of the capacitor as a function of time is given by:

$i=\frac{q_o}{RC}*e^{-\frac{t}{\tau}}\\where:\\\tau=RC\\$

$\tau=1.22*10^3*2.02*10^{-9}\\\tau=2.46*10^{-6}s$

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{9\µs}{2.46\µs}}\\\\i=0.048A$

$q=q_o*e^{-\frac{t}{\tau}}$

$q=4.55\µC*e^{-\frac{8\µs}{2.46\µs}}\\q=0.18\µC$

c) the maximum current occurs when t=0

$i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{0\µs}{2.46\µs}}\\\\i=1.85A$

6 0

3 years ago

PLEASE HELP WILL MARK BRAINLIEST

aniked [119]

2kg objects in the world of a hunting group

6 0

2 years ago