A joule is equivalent to a

What do we call fixed point around which a lever pivots?

xxTIMURxx [149]

Answer is A) Fulcrum

The fixed point that a lever rotates around is called the fulcrum.

4 0

2 years ago

The visible light portion of the electromagnetic spectrum is often subdivided into the colors of red, orange, yellow, green, blu

Ugo [173]

Light at the red end of the visible portion has the least energy, lowest frequency, same speed, and longer wavelength compared to the violet end.

<h3><u>Explanation:</u></h3>

The range in which the light exists is described as the electromagnetic spectrum. The light waves, radio waves, gamma rays,etc that exist in the world is not visible to human eyes. A kind of wave that modifies magnetic and electric fields is light. Spectroscopy makes use of all the frequencies and the wavelengths of the electromagnetic radiation.

The part of the electromagnetic spectrum that can be seen by the human eyes is the visible spectrum. The light waves with the wavelengths of 380 to 740 nm can be sen by the human eyes. Light at the red end of the visible portion has the least energy, lowest frequency, same speed, and longer wavelength compared to the violet end.

5 0

3 years ago

A projectile of mass 0.2 kg and an initial velocity of 50 m/s collides with the end of a blade attached to a turbine. The rotati

fgiga [73]

Answer:

5.5 rad/sec

Explanation:

8 0

3 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

3 years ago

A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W

maxonik [38]

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

5 0

3 years ago