What does the path of an object look like when it is in uniform motion ?

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

RideAnS [48]

Answer:

a) 2.33 m/s

b) 5.21 m/s

c) 882 m³

Explanation:

Using the concept of continuity equation

for flow through pipes

$A_{1}\times V_{1} = A_{2}\times V_{2}$

Where,

A = Area of cross-section

V = Velocity of fluid at the particular cross-section

given:

$A_{1} = 0.070 m^{2}$

$V_{1} = 3.50 m/s$

a) $A_{2} = 0.105 m^{2}$

substituting the values in the continuity equation, we get

$0.070\times 3.50 = 0.105\times V_{2}$

or

$V_{2} = \frac{0.070\times 3.5}{0.150}m/s$

or

$V_{2} = 2.33m/s$

b) $A_{2} = 0.047 m^{2}$

substituting the values in the continuity equation, we get

$0.070\times 3.50 = 0.047\times V_{2}$

or

$V_{2} = \frac{0.070\times 3.5}{0.047}m/s$

or

$V_{2} = 5.21m/s$

c) we have,

Discharge $Q = Area (A)\times Velocity(V)$

thus from the given value, we get

$Q = 0.070m^{2}\times 3.5m/s\$

$Q = 0.245 m^{3}/s$

Also,

Discharge $Q = \frac{volume}{time}$

given time = 1 hour = 1 ×3600 seconds

substituting the value of discharge and time in the above equation, we get

$0.245m^{3}/s = \frac{volume}{3600s}$

or

$0.245m^{3}/s\times 3600 = Volume$

volume of flow = $882 m^{3}$

8 0

3 years ago

A 2.74 g coin, which has zero potential energy at the surface, is dropped into a 12.2 m well. After the coin comes to a stop in

nata0808 [166]

Answer:

- 0.328J

Explanation:

POTENTIAL ENERGY = mgh

= 0.00274 × 9.81×12.2 here 2.74g = 0.00274kg

= 0.32792868 J

= 0.328J

AS IT IS BELOW THE SURFACE HENCE MUST BE NEGATIVE

hence potential energy = - 0.328J

8 0

3 years ago

Power is the time rate of change of work. a. True b. False

Bumek [7]

The statement about "Power is the time rate of change of work" is true. The answer is letter A. Power is presented by the equation P = Wt where P is the power, W is work and t is time.

6 0

3 years ago

Consider two planets in space that gravitationally attract each other. if the masses of both planets are doubled, and the distan

amid [387]

Yhuihoifjhh F = Gm1m2 / r^2

if the masses are doubled then the force is increased by a factor of 4

if the distance is doubled the force is decreased by a factor of 1/ 2^2
the net result is no change in force

6 0

3 years ago

Read 2 more answers

A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x

makkiz [27]

Answer:

A)Magnitude of the net electric field at the origin due q₁ and q₂

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

β=3.91°

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1nC= 10⁻⁹ C

Data

q₁ = -4 nC = -4*10⁻⁹ C

q₂ = +6nC =+ 6*10⁻⁹ C

k = 9*10⁹ N*m²/C²

$d_{1} =\sqrt{0.6^{2}+0.8^{2} } =1 m$

d₂ = 0.6 m

Graphic attached

The attached graph shows the field in x=0, y=0, due to the charges q₁ and q₂:

E₁: Total field at point x=0 , y=0 due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge.

E₂: Total field at point x=0 , y=0 due to charge q₂. As the charge q2 is positive (q₂+) ,the field leaves the charge.

Calculation of the electric field at the origin of x-y coordinates due to the charge q₁

E₁=K*q₁/d₁²=9*10⁹*4*10⁻⁹/1²=36N/C

Components (x-y) of the field due to q1:

E₁x=E₁*cosα = 36*(0.6/1) = 36*(0.6) = 21.6 N/C

E₁y=E₁*sinα = 36*(0.8/1) = 36*(0.8) =28.8 N/C

Calculation of the electric field at the origin of x-y coordinates due to the charge q₂

E₂=K*q₂/d₂² = -9*10⁹*6*10⁻⁹/0.6² = -150 N/C

Calculation of the electric field components at the origin of x-y coordinates

E₀ is the electric field at the origin of x-y coordinates (x=0,y=0)

E₀x= E₁x+E₂= 21.6-150 = -128.4 N/C

E₀y= E₁y = 28.8 N/C

A )Magnitude of the net electric field at the origin due q₁ and q₂

$E_{o} =\sqrt{128.4^{2}+28.8^{2} }$

E₀= 131.6 N/C

B) Direction of the net electric field at the origin due q₁ and q₂

$\beta =tanx{-1} \frac{E_{oy} }{Eox}$

$\beta =tan^{-1} \frac{28.8}{128.4}$

β=3.91°

6 0

4 years ago