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Digiron [165]
3 years ago
6

Magnetic fields go from the north poles of magnets to the south poles of magnets. true or false

Physics
1 answer:
Novay_Z [31]3 years ago
7 0

The Answer is true, the saying opposites attract are true for magnets when poles match though they repel.

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Match each example with the appropriate stage of technological design. drag each arrow to its correct match.
Akimi4 [234]

Answer:

Review each answer in the explanation part

Explanation:

Identify a problem or need = Aerospace engineers need a lightweight material to build a jet

Note: Identify a problem is related always with a need.

After identify the problem the next stage is the design

Design a solution = Engineers determine what features the material must have.

Note: The design process implies the choice of the materials and the proper numeric calculations.

Implement the solution = Engineers test a carbon - plastic compound in a wind tunnel.

Note: after the design, the following process is the built of the equipment and the different tests.

Evaluate the solution = The carbon-plastic compound is redesigned to save maximun energy.

Note: After testing, you enter a part of re-design that means making improvements to the already constructed.

7 0
3 years ago
Someone pls answer this with steps
Ivanshal [37]
No worry’s I am here to help
8 0
3 years ago
The circuit you should use to find the open-circuit voltage is
fiasKO [112]

Answer:

Incomplete questions check attachment for circuit diagram.

Explanation:

We are going to use superposition

So, we will first open circuit the current source and find the voltage Voc.

So, check attachment for open circuit diagram.

From the diagram

We notice that R3 is in series with R4, so its equivalent is given below

Req(3-4) = R3 + R4

R(34) = 20+40 = 60 kΩ

Notice that R2 is parallel to the equivalent of R3 and R4, then, the equivalent of all this three resistor is

Req(2-3-4) = R2•R(34)/(R2+R(34))

R(234) = (100×60)/(100+60)

R(234) = 37.5 kΩ

We notice that R1 and R(234) are in series, then, we can apply voltage divider rule to find voltage in R(234)

Therefore

V(234) = R(234) / [R1 + R(234)] × V

V(234) = 37.5/(25+37.5) × 100

V(234) = 37.5/62.5 × 100

V(234) = 60V.

Note, this is the voltage in resistor R2, R3 and R4.

Note that, R2 is parallel to R3 and R4. Parallel resistor have the same voltage, then voltage across R2 equals voltage across R34

V(34) = 60V.

Now, we also know that R3 and R4 are in series,

So we can know the voltage across R4 which is the Voc we are looking for.

Using voltage divider

V4 = Voc = R4/(R4 + R(34)) × V(34)

Voc = 40/(40+60) × 60

Voc = 24V

This is the open circuit Voltage

Now, finding the short circuit voltage when we short circuit the voltage source

Check attachment for circuit diagram.

From the circuit we notice that R1 and R2 are in parallel, so it's equivalent becomes

Req(1-2) = R1•R2/(R1+R2)

R(12) = 25×100/(25+100)

R(12) = 20 kΩ

We also notice that the equivalent of Resistor R1 and R2 is in series to R3. Then, the equivalent resistance of the three resistor is

Req(1-2-3) = R(12) + R(3)

R(123) = 20 + 20

R(123) = 40 kΩ

We notice that, the equivalent resistance of the resistor R1, R2, and R3 is in series to resistor R4.

So using current divider rule to find the current in resistor R4.

I(4) = R(123) / [R4+R(123)] × I

I(4) = 40/(40+40) × 8

I(4) = 4mA

Then, using ohms law, we can find the voltage across the resistor 4 and the voltage is the required Voc

V = IR

V4 = Voc = I4 × R4

Voc = 4×10^-3 × 40×10^3

Voc = 160V

Then, the sum of the short circuit voltage and the open circuit voltage will give the required Voc

Voc = Voc(open circuit) + Voc(short circuit)

Voc = 24 + 160

Voc = 184V.

3 0
3 years ago
A computer base unit of mass 7.5 kg is dragged along a smooth desk. If the normal contact force is 23N and the tension in the ar
kramer
<span>If there isn't any force then the normal contact force will be 


N=m*g=7.5*9.81=73.58N 

which is 73.58-23=50.58N less 

so, there the person must pull at 23 degree upward 

break down the tension in two components, vertical and horizontal. 


vertical tension= 50.58=T*sin23 

T=50.58/sin23=129.45N</span>
7 0
3 years ago
Read 2 more answers
The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
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