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3 years ago

Match the chemical formula to the correct name.

Physics

1 answer:

11Alexandr11 [23.1K]3 years ago

3 0

Co carbon monoxide
sorry friend i don't know other ones

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A force of 333 N is applied 33 cm from the pivot point. What is the maximum torque of this situation

exis [7]

Answer: 109.89 Nm

Explanation:

The maximum torque will be calculated as the force multiplied by the perpendicular distance. This will be:

Torque = force × perpendicular distance

torque = 333 × 0.33

= 109.89 Nm

5 0

3 years ago

A 21.3 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.7 cm from

Bezzdna [24]

Answer:

The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.

Explanation:

Given;

electric current, I = 21.3 A

distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m

The strength of the resulting magnetic field at the given distance is calculated as;

$B = \frac{\mu_o I}{2\pi R}$

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A

$B = \frac{\mu_o I}{2\pi R}\\\\B = \frac{4\pi*10^{-7} *21.3}{2\pi(0.457)}\\\\B = 9.322 *10^{-6} \ T$

Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.

7 0

4 years ago

A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular

Fynjy0 [20]

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

3 0

3 years ago

A single beam of light is shone into a glass containing milky water.

Bumek [7]

Answer:

the first one

Explanation:

3 0

3 years ago

What visible wavelengths of light are strongly reflected from a 390-nm-thick soap bubble?

DiKsa [7]

Answer:

So visible wavelength which is possible here is

416 nm and 693.3 nm

Explanation:

As we know that for normal incidence of light the path difference of the reflected ray is given as

$2\mu t + \frac{\lambda}{2} = \Delta x$

so here we can say that for maximum intensity condition we will have

$\Delta x = N\lambda$

so we have

$2\mu t + \frac{\lambda}{2} = N\lambda$

now for visible wavelength we have

for N = 1

$2\mu t = \frac{\lambda}{2}$

$\lambda = 4\mu t$

$\lambda = 4(\frac{4}{3})(390 nm)$

$\lambda = 2080 nm$

for N = 2

$\lambda = \frac{4\mu t}{3}$

$\lambda = \frac{4(\frac{4}{3})(390 nm)}{3}$

$\lambda = 693.3 nm$

for N = 3

$\lambda = \frac{4\mu t}{5}$

$\lambda = \frac{4(\frac{4}{3})(390 nm)}{5}$

$\lambda = 416 nm$

6 0

4 years ago