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4 years ago

A water pump is a positive displacement-type pump true or false

Physics

1 answer:

Kay [80]4 years ago

6 0

Answer: True

A water pump belong to a positive displacement pump that provides constant flow of water at fixed speed, regardless of changes in the counter pressure. The two main types of positive displacement pump are rotary pumps and reciprocating pumps.

Moreover, water pump is a reciprocating positive displacement pump that have an expanding cavity on the suction side and a decreasing cavity on the discharge side. In water pumps, the liquid flows into the pumps as the cavity on the suction side expands and then the liquid flows out of the discharge as the cavity collapses providing water in a pail.

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Which is not part of nuclear chemistry? A. Strong nuclear force B. Double-replacement reactions C. Radioactivity D. Nuclear deca

ziro4ka [17]

The correct choice is

B. Double-replacement reactions

in this type of reaction , two ionic compounds react and swaps their cations or anions giving two new compounds in the product. hence the double replacement reactions does not deal with the nucleus. all other terms like strong nuclear forces, radioactivity and nuclear decay deal with the nucleus.

8 0

4 years ago

Read 2 more answers

A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe

ikadub [295]

Answer:

$f_{fr}=1590.85 N$

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

$f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r}$ (1)

f(fr) is the friction force
N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

$Ncos(25)-mg-f_{fr}sin(25)=0$ (2)

Let's combine (1) and (2) to find f(fr)

$f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}$

$f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}$

$f_{fr}=1590.85 N$

I hope it helps you!

5 0

3 years ago

An apple falls out of a tree from a height of 2.3 m What is the impact speed of the apple?

ivann1987 [24]

Answer:

6.72 m/s

Explanation:

recall that the equations of motion may be expressed as

v² = u² + 2as

where,

v = final velocity,

u = initial velocity = 0 m/s because it is stationary before it starts falling

a = acceleration (in this case because it is falling, it is the acceleration due to gravity = 9.81 m/s²)

s = distance traveled = 2.3m

in our case, if we neglect air resistance, then we simply substitute the known values above into the equation of motion.

v² = u² + 2as

v² = 0² + 2(9.81)(2.3)

v² = 45.126

v = √45.126

v = 6.72 m/s

6 0

4 years ago

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Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

4 years ago

What happens to a car that experiences an increase in kinetic energy

Allisa [31]

Answer:

Your moving vehicle has kinetic energy; as you increase your vehicle's speed, your vehicle's kinetic energy increases. The greater your vehicle's kinetic energy, the greater the effort that will be required to stop the vehicle.

Explanation:

I hope this explained up above. Have a great dayyyy:)

4 0

3 years ago

Read 2 more answers