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Otrada [13]
3 years ago
12

After brainstorming, engineers may refine their ideas by making what???

Physics
1 answer:
denpristay [2]3 years ago
5 0
Engineers refine their ideas by making sketches and models.
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You can use one of your 5 senses to make _____ during an inquiry activity.
aivan3 [116]

you can use one of your 5 senses to make _____ during an inquiry activity.

sound

4 0
4 years ago
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Why do electrons transfer charge and not protons?
mel-nik [20]
<h2>Answer:Protons interact in ways that electrons do not. ... Electrons are not affected by the strong force, and so they only get trapped by the electrical attraction to the nucleus which is much weaker in ionized atoms.</h2><h2 /><h2>Explanation:Therefore it is easier for electrons to move away from one atom to another, transferring charge.</h2>

3 0
3 years ago
What temperature resistivity of tungsten is 4 times resistivity of silver?
mihalych1998 [28]

Answer:

Slightly above 20°c

Explanation:

The resistivity of silver is 1.59 x 10-8Ωm{ as discovered by experiment from scientist}

The resistivity of tungsten at 20°c is 5.6x10-8Ωm

We see the value is just an approximation of 4 times not exactly.

6 0
3 years ago
An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef
Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field B=5.0\times10^{-5}\ T

(a). We need to calculate the induced emf

Using formula of emf

E = Blv\sin\theta

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}

E=1.398\times10^{-3}\ V

We need to calculate the induced current

E =IR

I=\dfrac{E}{R}

Put the value into the formula

I=\dfrac{1.398\times10^{-3}}{3.0}

I=0.5\ mA

(b). We need to calculate the power dissipated through the resistor

Using formula of power

P=I^2 R

Put the value into the formula

P=(0.5\times10^{-3})^2\times3.0

P=7.5\times10^{-7}\ Watt

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

6 0
3 years ago
Read 2 more answers
The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoi
Novay_Z [31]

Answer: 0.43 V

Explanation:

L = [μ(0) * N² * A] / l

Where

L = Inductance of the solenoid

N = the number of turns in the solenoid

A = cross sectional area of the solenoid

l = length of the solenoid

7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24

1.752*10^-3 = 4π*10^-7 * 202500 * A

1.752*10^-3 = 0.255 * A

A = 1.752*10^-3 / 0.255

A = 0.00687 m²

A = 6.87*10^-3 m²

emf = -N(ΔΦ/Δt).........1

L = N(ΔΦ/ΔI) so that,

N*ΔΦ = ΔI*L

Substituting this in eqn 1, we have

emf = - ΔI*L / Δt

emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3

emf = 0.0234 / 0.055

emf = 0.43 V

6 0
3 years ago
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