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cluponka [151]
3 years ago
9

Water flows with constant speed through a garden hose that goes up to 27.5 cm high. if the water pressure is 132kpa at the botto

m of the step, what is its pressure at the top of the step?
Physics
1 answer:
sergejj [24]3 years ago
6 0

Answer:

The pressure at the top of the step is 129.303 kilopascals.

Explanation:

From Hydrostatics we find that the pressure difference between extremes of the water column is defined by the following formula, which is a particular case of the Bernoulli's Principle (v_{bottom}\approx v_{top}):

p_{bottom}-p_{top} = \rho\cdot g\cdot \Delta h (1)

p_{bottom}, p_{top} - Total pressures at the bottom and at the top, measured in pascals.

\rho - Density of the water, measured in kilograms per cubic meter.

\Delta h - Height difference of the step, measured in meters.

If we know that p_{bottom} = 132000\,Pa, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} and \Delta h = 0.275\,m, then the pressure at the top of the step is:

p_{top} = p_{bottom}-\rho\cdot g\cdot \Delta h

p_{top} = 132000\,Pa-\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.275\,m)

p_{top} = 129303.075\,Pa

p_{top} = 129.303\,kPa

The pressure at the top of the step is 129.303 kilopascals.

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3 years ago
A block of mass m=16.8 kg is sliding on a surface with initial velocity v=23.2 m/s. The block has a coefficient of kinetic frict
Firdavs [7]

Answer:

t = 23.92 s

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block  and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction, downward

FN : Normal force : perpendicular to the floor, upward

fk :  Kinetic friction force: parallel to the floor  and opposite to the movement

F = 86.4 N , in the direction of the motion

Calculated of the W

W= m*g

W=  16.8 kg* 9.8 m/s² = 164.64 N

Calculated of the FN  

We apply the formula (1)  

∑Fy = m*ay ay = 0  

FN - W = 0  

FN = W  

FN =  164.64 N

Calculated of the fk

fk  = μk*FN

fk  = 0.426* 164.64 N

fk  = 70.13 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

F - fk = m*a

86.4 -70.13  = (16.8)*(-a)

16.26 =  (16.8)*(-a)

a = -(16.26 )/ (16.8)

a = - 0.97 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula :

vf = v₀ + a*t   Formula (2)

Where:  

t: time interval  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 23.2 m/s

vf = 0

a =  -0.97 m/s²  

Time it takes for the block to stop

We replace data in the formula (2)  to calculate the time

vf= v₀+a*t

0 = 23.2+( -0.97)*t

(0.97)*t  = 23.2

(0.97)*t  = 23.2

t = 23.2 / (0.97)

t = 23.92 s

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Before the fission process takes place, lead-207 is bombarded with neutrons, it can change into
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Lead-207 can change into Lead-208.

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If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0
3 years ago
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