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tensa zangetsu [6.8K]
3 years ago
10

A filled water tower sits on the top of the highest hill near a city. The cylindrical tower has a height of 55.0 m

Physics
1 answer:
Reika [66]3 years ago
7 0

Answer:

539 kPa

Explanation:

Pressure equals density times acceleration of gravity times depth.

P = ρgh

Water has a density of 1000 kg/m³, and acceleration of gravity is 9.8 m/s².

P = (1000 kg/m³) (9.8 m/s²) (55.0 m)

P = 539,000 Pa

P = 539 kPa

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What potential difference would an electron have to fall through to acquire a speed of 3.00*10^6 m/sec?
kirill [66]

Answer:

25.6 V

Explanation:

The kinetic energy of electron associated with its potential difference is given by eV which is equal to the 1/2 mv^2.

m = 9.1 x 10^-31 kg, v = 3 x 10^6 m/s, e = 1.6 x 10^-19 C

eV = 1/2 m v^2

V = mv^2 / 2 e

V = (9.1 x 10^-31) x (3 x 10^6)^2 / (2 x 1.6 x 10^-19)

V = 25.6 V

4 0
3 years ago
Classical physics is a good approximation to modern physics under certain circumstances. What are they?
Aleks [24]

Answer and Explanation:

In classical physics we study about macroscopic objects and in modern physics we study about microscopic objects but under some circumstances classical physics is a approximation to modern physics such as

  • The object must be microscopic but they should be large enough so that we can see that through a microscope
  • The speed of the object should be very less it should be about 1 % of the speed of the light
  • They should be involved with very week gravitational field

4 0
3 years ago
The sensor in the torso of a crash test dummy records the magnitude znd direction of the net force acting on the dummy.If the du
muminat

Here in the given situation two forces are acting on the dummy from two perpendicular directions

Since force is a vector quantity so we need to add two or more forces by vector addition method

Now here we know that two forces are perpendicular to each other

So we will use Pythagoras theorem

F_{net} = \sqrt{F_1^2 + F_2^2}

given that

F_1 = 130.0 N

F_2 = 4500.0 N

now plug in all in the above equation

F_{net} = \sqrt{130^2 + 4500^2}

F_{net} = 4501.9 N

So the sensor will report net force as 4501.9 N

7 0
3 years ago
A solid sphere rolls along a horizontal, smooth surface at a constant linear speed without slipping. What is the ratio between t
Pavlova-9 [17]

The ratio of the rotational kinetic energy about the center of the sphere and the total kinetic energy of the sphere is 2/7.

<h3>What is the rotational kinetic energy of a sphere?</h3>

The rotational kinetic energy of the sphere is directly proportional to the square of angular acceleration.

K = \dfrac 12 I\omega ^2

Where,

I = rotational inertia of sphere = 2/5MR²

Where M is the mass of the sphere and R is the radius of the sphere,

ω = angular acceleration of sphere = V/R

Where V is the speed of the sphere.

So,

K = 1/2 × 2/5MR² × V²/R²

K = MV²/5

The translational kinetic energy of the sphere,

K' = \dfrac 12 MV^2

The total kinetic energy of the sphere,

K_t = K + K'

So,

K_t= \dfrac {MV^2}5 +\dfrac { MV^2}2\\\\K_t= \dfrac {7MV^2}{10}

Thus the Ratio of rotational kinetic energy to total kinetic energy,

\dfrac KK_t = \dfrac {\dfrac {MV^2}{5} }{ \dfrac {7MV^2}{10}}\\\\\dfrac KK_t = \dfrac 15 \times \dfrac { 10}7\\\\\dfrac KK_t= \dfrac 27

Therefore, the ratio of the rotational kinetic energy about the center of the sphere and the total kinetic energy of the sphere is 2/7.

Learn more about rotational kinetic energy:

brainly.com/question/14611300

3 0
2 years ago
The pressure difference, , across a partial blockage in an artery (called a stenosis) is approximated by the equation where is t
Strike441 [17]

The question is incomplete. The complete question is  :

The pressure difference, Δp, acK_uross a partial blockage in an artery (called a stenosis) is approximated by the equation :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Where V is the blood velocity, μ the blood viscosity {FT/L2}, ρ the blood density {M/L3}, D the  artery diameter, A_0 the area of the unobstructed artery, and A1 the area of the stenosis.  Determine the dimensions of the constants K_v and K_u. Would this equation be valid in any  system of units?

Solution :

From the dimension homogeneity, we require :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Here, x means dimension of x. i.e.

$[ML^{-1}T^{-2}]=\frac{[K_v][ML^{-1}T^{-1}][LT^{-1}]}{[L]}+[K_u][1][ML^{-3}][L^2T^{-2}]$

                    $=[K_v][ML^{-1}T^{-2}]+[K_u][ML^{-1}T^{-2}]$

So, $[K_u]=[K_v]=[1 ]=$ dimensionless

So, K_u and K_v  are dimensionless constants.

This equation will be working in any system of units. The constants K_u and K_v will be different for different system of units.

5 0
3 years ago
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