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3 years ago

What are two factors affecting friction? How do they affect friction? ch

Physics

1 answer:

AlexFokin [52]3 years ago

5 0

The roughness of the surface, the mass of the object, and the area of contact.

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A sound that is perceived as twice as loud requires

iren [92.7K]

Answer:

C.ten times the intensity.

7 0

3 years ago

Monochromatic light passes through a double slit, producing interference, the distance between the slit centres is 1.2 mm and th

Alik [6]

Answer:

The wavelength of the light is $7200\ \AA$ .

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes $\beta= 0.3\ cm$

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

$\beta=\dfrac{D\lambda}{d}$

Put the value into the formula

$0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}$

$\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}$

$\lambda=7.2\times10^{-7}\ m$

$\lambda=7200\ \AA$

Hence, The wavelength of the light is $7200\ \AA$ .

8 0

3 years ago

Three-fourths of the area of a rectangular lawn 30 feet wide by 40 feet long is to be enclosed by a rectangular fence. If the en

satela [25.4K]

Answer:

The fence is 5feet less.

Explanation:

We need to determine

The less amount of fence required, if the enclosure has full width and reduced length, compared to full length and reduced width.

Approach & WorkingArea of lawn = 30 × 403/4th of the area of lawn = ¾(30 × 40) = 30 * 30

When full width will be fenced, and reduced length will be fenced.

Width = 30 feet30 * L = 30 * 30Hence, length = 30 feetLength of fence needed = 2(30 + 30) = 120 feet

When full length will be fenced, and reduced width will be fenced

Length = 40 feet40 * W = 30 * 30W = 22.5 feetLength of fence needed = 2(40 + 22.5) = 125 feet

Difference in length of fence needed = 125 – 120 = 5 feet.

4 0

4 years ago

The opening to a cave is a tall, 30.0-cm-wide crack. A bat is preparing to leave the cave emits a 30.0 kHz ultrasonic chirp. How

Vlada [557]

Answer:

The value is $w = 7.54 \ m$

Explanation:

From the question we are told that

The length of the crack is $a = 0.3 \ m$

The frequency is $f = 30.0 \ kHz = 30 *10^{3} \ Hz$

The distance outside the cave that is being consider is $D = 100 \ m$

The speed of sound is $v_s = 340 \ m/s$

Generally the wavelength of the wave is mathematically represented as

$\lambda = \frac{v}f}$

=> $\lambda = \frac{340 }{30*10^{3}}$

=> $\lambda = 0.0113 \ m/s$

Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as

$y = \frac{ n * \lambda * D}{a}$

=> $y = \frac{ 1 * 0.0113 * 100}{0.3}$

=> $y = 3.77 \ m$

Generally the width of the sound beam is mathematically represented as

$w = 2 * y$

=> $w = 2 * 3.77$

=> $w = 7.54 \ m$

4 0

3 years ago

What types of cells are likely to have lots of smooth ER?

Andrei [34K]

Answer:

eukaryotic cells

Explanation:

"Smooth endoplasmic reticulum (sER) is (a part of) endoplasmic reticulum that is tubular in form and lacks ribosomes. It is present in eukaryotic cells and is associated with lipid synthesis, carbohydrate metabolism, regulation of calcium concentration, and drug detoxification"

source: biologyonline

8 0

3 years ago