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Pepsi [2]
3 years ago
14

What are electric motors?

Physics
1 answer:
Lapatulllka [165]3 years ago
7 0
It's an electric motor that converts electric energy into Mechanical energy
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A pump uses a piston of 15 cm diameter that moves at 2.0 cm/s as it pushes a fluid through a pipe. what is the speed of the flui
Nitella [24]
The important point here is that volumetric flow rate in the pump and the pipe is the same.

Q = AV, where Q = Volumetric flow  rate, A = Cross sectional area, V = velocity

Q (pump) = (π*15^2)/4*2 = 353.43 cm^3/s
Q (pipe) = (π*(3/10)^2)/4*V = 0.071V

Q (pump) = Q (pipe)
0.071V = 353.43 => V = 5000 cm/s

Therefore, the flow  of water in the pipe is 5000 cm/s.
4 0
3 years ago
A piece of wood has a volume of 4.6 cm to the power of 3, and a mass of 3.7 g what is the density of the wood
AysviL [449]

Density is given by:

D = M/V

D = density, M = mass, V = volume

Given values:

M = 3.7g, V = 4.6cm³

Plug in and solve for D:

D = 3.7/4.6

D = 0.80g/cm³

3 0
3 years ago
What is the pendulum length whose period is 2.0s ?
Mashutka [201]
Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24mT=2 \pi   \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi }  = \frac{L}{g} \\
 \frac{T^2}{2 \pi }*g=L\\
L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m

7 0
3 years ago
The plain flight covers 400 miles in the first hour. As you approach the landing site, it takes you an hour to cover the last 20
tatyana61 [14]

that is an example of negative acceleration because it is slowing down

3 0
3 years ago
two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y
svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

v_0_x=6380 m/s

v_0_y=6770 m/s

time period of engine running=763 s

Displacement in x=4.50\times 10^6

y=7.27\times 10^6

Using s=ut+\frac{at^2}{2} in x and y direction

x=v_0_x\times t+\frac{at^2}{2}

4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}

4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}

a=-1.23 m/s^2

In y direction

y=v_0_y\times t+\frac{a't^2}{2}

7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}

7.27\times 10^6-5.16\times 10^6=\frac{a\times 763^2}{2}

a=7.24 m/s^2

x component=-1.23 m/s^2

y component=7.24 m/s^2

3 0
3 years ago
Read 2 more answers
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