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Sloan [31]
2 years ago
14

A conveyer belt moves a 40 kg box at a velocity of 2 m/s. What is the kinetic energy of the box while it is on the conveyor belt

?
Physics
2 answers:
andrew-mc [135]2 years ago
4 0
Kinetic energy =  \frac{mass*velocity ^{2} }{2} =  \frac{40*2 ^{2} }{2} = 80J. Therefore kinetic energy of the box while on the conveyor belt is 80J. 
dlinn [17]2 years ago
4 0

Answer : The kinetic energy of the box is, 80 J

Explanation :

The formula used for kinetic energy is:

K.E=\frac{1}{2}mv^2

where,

K.E = kinetic energy = ?

m = mass = 40 kg

v = velocity = 2 m/s

Now put all the given values in the above formula, we get the kinetic energy of the box while it is on the conveyor belt.

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}\times (40kg)\times (2m/s)^2

K.E=80kg.m^2/s^2=80J

Thus, the kinetic energy of the box is, 80 J

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2 years ago
Charge: A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have? (
snow_lady [41]

Answer

given,

net charge = +2.00 μC

we know,

1 coulomb charge =  6.28 x 10¹⁸electrons

1 micro coulomb  charge =  6.28 x 10¹⁸ x 10⁻⁶ electron

                                         = 6.28 x 10¹² electrons

2.00 μC = 2 x 6.28 x 10¹² electrons

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since net charge is positive.

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hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.

6 0
2 years ago
A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at
GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

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Angelina_Jolie [31]
Hello.

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Now that we have the value for Work, let's apply it to our Power formula.
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The Power required to lift the girder is 1944.44~ W (Unit for Power is "W" or "Watts").

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