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AVprozaik [17]
3 years ago
7

Which of the following items involve a wedge?

Physics
2 answers:
Romashka-Z-Leto [24]3 years ago
6 0
B because under the blade it curves
Fynjy0 [20]3 years ago
6 0
I did some research and the answers are A. a chisel B. an axe and C. a knife
You might be interested in
In this diagram of the water cycle, clouds form in the air due to what process? A) run off. B) condensation. C) transpiration. D
Fiesta28 [93]
I can't see the diagram, but it would be condensation. (You can use the process of elimination here as well: precipitation is when rain/snow/sleet/hail falls, run off is.. well, run off, and transpiration is about trees)
4 0
4 years ago
Read 2 more answers
A grating whose slits are 3.2x10^-4 cm apart produces a third-order fringe at a 25.°0 angle. What is the wavelength of light tha
Ber [7]

Answer:

The light used has a wavelenght of 4.51×10^-7 m.

Explanation:

let:

n be the order fringe

Ф be the angle that the light makes

d is the slit spacing of the grating

λ be the wavelength of the light

then, by Bragg's law:

n×λ = d×sin(Ф)

λ = d×sin(Ф)/n

λ = (3.2×10^-4 cm)×sin(25.0°)/3

  = 4.51×10^-5 cm

  ≈ 4.51×10^-7 m

Therefore, the light used has a wavelenght of 4.51×10^-7 m.

7 0
3 years ago
Which action would be most important to perform in a controlled experiment? ___A. Design an experiment that is based on previous
ludmilkaskok [199]

Answer:

D. Test a piece of treated iron and a piece of untreated iron under the same conditions.

Explanation:

A controlled experiment is defined as an experiment where the experimenter or the researcher tries to test or experiment his result by changing one variable at a time and keeping the other factors under the same conditions and sees how it effects the result.

The factors which can be changed in an experiment is called its variables.

In the context, testing a piece of treated iron as well as a untreated piece of iron under the same working conditions forms a controlled experiment.

4 0
3 years ago
The induced magnetic field at radial distance 4.0 mm from the central axis of a circular parallel-plate capacitor is 1.8 ✕ 10−7
wel

Answer:

\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}

Explanation:

According to Gauss's law, the electric flux through the circular plates is defined as the electric field multiplied by its area:

\Phi=EA=E(\pi R^2)(1)

The magnetic field around the varying electric field of the circular plates is given by:

B=\frac{\epsilon_0 \mu_o}{2\pi r}\frac{d\Phi}{dt}(2)

Replacing (1) in (2) and solving for \frac{dE}{dt}:

B=\frac{\epsilon_0 \mu_o\pi R^2}{2\pi r}\frac{dE}{dt}\\\frac{dE}{dt}=\frac{2rB}{\epsilon_0 \mu_o R^2}\\\frac{dE}{dt}=\frac{2(4*10^{-3}m)(1.8*10^{-7}T)}{(8.85*10^{-12}\frac{C^2}{N\cdot m^2})(4\pi *10{-7}\frac{Tm}{A})(2.5*10^{-3}m)^2}\\\\\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}

3 0
3 years ago
Looking at group 2 on the periodic table which of these elements probably has the lowest ionic radius
lakkis [162]
I think the answer is Helium.

Please tell me if I'm wrong.
7 0
3 years ago
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