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3 years ago

HELP ME PLEASE ILL MAKE U BRANLEIST IF WRITE

Physics

2 answers:

miss Akunina [59]3 years ago

7 0

Left left right im pretty sure

katen-ka-za [31]3 years ago

3 0

so the 1st on is the one on the left, middle is right and the 3rd one is the right one

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Which is true?A)The spin angular momentum vector itself can be measured, and it can be aligned along the measurement axis (z axi

12345 [234]

Answer:

Option A is correct

Explanation:

Direction of spin can be measured because of the behaviours of the particle as it goes through a magnetic field and the size of the spin is dependant on type of particle.

3 0

3 years ago

it takes a school bus full of kids longer to stop than it does a small car with a single passenger. which law is this an example

exis [7]

Newton´s second law of motion

7 0

4 years ago

Compare the light gathering power of a 1 meter diameter telescope to that of the human eye ,which has a diameter of roughly 2.5

lesantik [10]

Answer:

The telescope can gather light 1600 times more than the human eyes can!

Explanation:

The light gathering ability of an optical element is directly proportional to its area of opening.

So, in comparing the light gathering abilities for two objects, it is just the ratio of their area of opening.

Let the diameter of the telescope be D = 1 m

And the diameter of the human eyes be d = 2.5 cm = 0.025 m

Light gathering ability of the telescope compared to the eyes = D² ÷ d²

= (D²/d²) = (1²/0.025²) = 1600 times.

The telescope can gather light 1600 times more than the human eyes can!

Hope this Helps!!!

7 0

4 years ago

A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center

uranmaximum [27]

Answer:

4.36 rad/s

Explanation:

Radius of platform r = 2.97 m

rotational inertia I = 358 kg·m^2

Initial angular speed w = 1.96 rad/s

Mass of student m = 69.5 kg

Rotational inertia of student at the rim = mr^2 = 69.5 x 2.97^2 = 613.05 kg.m^2

Therefore initial rotational momentum of system = w( Ip + Is)

= 1.96 x (358 + 613.05)

= 1903.258 kg.rad.m^2/s

When she walks to a radius of 1.06 m

I = mr^2 = 69.5 x 1.06^2 = 78.09 kg·m^2

Rotational momentuem of system = w(358 + 78.09) = 436.09w

Due to conservation of momentum, we equate both momenta

436.09w = 1903.258

w = 4.36 rad/s

7 0

3 years ago

Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is

Elan Coil [88]

Answer:

The intensity of light from the 1mm from the central maximu is $I = 0.822I_o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 620 nm = 620 *10^{-9}m$

The width of the slit is $w = 0.450mm = \frac{0.45}{1000} = 0.45*10^{-3} m$

The distance from the screen is $D = 3.00m$

The intensity at the central maximum is $I_o$

The distance from the central maximum is $d_1 = 1.00mm = \frac{1}{1000} = 1.0*10^{-3}m$

Let z be the the distance of a point with intensity I from central maximum

Then we can represent this intensity as

$I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2$

Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

$sin \theta = \frac{z}{D}$

if the angle between the the light at z and the central maximum is small

Then $sin \theta = \theta$

Which implies that

$\theta = \frac{z}{D}$

substituting this into the equation for the intensity

$I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D} ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ]$

given that $z =1mm = 1*10^{-3}m$

We have that

$I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2$

$=I_o [\frac{sin(0.760)}{0.760}] ^2$

$I = 0.822I_o$

4 0

3 years ago