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3 years ago

175 in expanded form

Physics

1 answer:

hjlf3 years ago

4 0

Answer:

In expanded notation the answer is 100,70 5

Explanation:

Hope this helps

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Study the interaction of the light rays with the mediums. Classify the medium in each image as being transparent or opaque.

Tresset [83]

Answer:

a) transparent

b) transparent

c) opaque

Explanation:

In the first one, the light rays go completely through, so it is transparent.

The second one I'm not too sure about. It is refraction so it's going through a different material, but the fact that it went through makes me say transparent.

Last one, the light rays are reflecting off the surface so it's opaque.

Please feel free to correct me if I'm wrong. This is just my understanding

6 0

3 years ago

Read 2 more answers

A stagehand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N. W

Diano4ka-milaya [45]

<u>Answer:</u>

Option (a)

<u>Explanation :</u>

A stage hand starts sliding a large piece of stage scenery originally at rest by pulling it horizontally with a force of 176 N.

Hence Force applied $\text { Fapplied }=176 \mathrm{N}$

Force on piece of scenery $\mathrm{F}_{\mathrm{g}}=490 \mathrm{N}$

$\Sigma \mathrm{F} \mathrm{y}=\mathrm{F} \mathrm{n}-\mathrm{Fg}=0$

$\mathrm{Fn}=\mathrm{Fg}$

$\mathrm{FS}_{\mathrm{max}}=\mathrm{F}_{\mathrm{applied}}$

µk = $\frac{\mathrm{Fs} \max }{\mathrm{Fn}}$

= $\frac{\text { Fapplied }}{\mathrm{Fg}}$

= $\frac{176}{490}$ =0.36

coefficient of static friction is 0.36

6 0

4 years ago

One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra

sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration ( $a_{r}$ ), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

$a_{r} = 4\pi^{2}\cdot f^{2}\cdot R$ (1)

Where:

$f$ - Frequency, measured in hertz.

$R$ - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension ( $T$ ), measured in newtons, through the string, then we derive the following model:

$\Sigma F = T = m\cdot a_{r}$ (2)

Where $m$ is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

$T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R$ (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

$\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right)$ (4)