Degree day calculations are based on a temperature of?

A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 610 N horizontal force to set it in motion. After the clo

denpristay [2]

Answer:0.669

Explanation:

Given

mass of clock 93 kg

Initial force required to move it 610 N

After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity

Initially static friction is acting which is more than kinetic friction

thus 613 force is required to overcome static friction

$\mu _smg=610$

$\mu _s\times 93\times 9.8=610$

$\mu _s=0.669$

5 0

3 years ago

A laser is shone through a double slit and a particular interference pattern is observed on a screen some distance away. If the

Grace [21]

If the separation between the openings in a laser is increased, then the distance between the interference fringes decreases

<h3>What is Interference fringe ?</h3>

Interference fringe refers to bands caused by different lights which can be found in phase or not each other.

Distances between laser fringes are short which is due to light wavelength.
The interference fringes can be estimated by knowing slit separation and wavelength.

In conclusion, if the separation between the openings in a laser is increased, then the distance between the interference fringe decreases

Learn more about Interference fringe here:

brainly.com/question/14264436

#SPJ4

5 0

2 years ago

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact wit

Stels [109]

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, $P=2.2\times 10^5\ Pa$

Area of each tire, $A=0.023\ m^2$

Area of 4 tires, $A=0.092\ m^2$

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

$P=\dfrac{F}{A}$

$F=P\times A$

$F=2.2\times 10^5\times 0.092$

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

7 0

3 years ago

A student measures the speed of yellow light in water to be 2.00x10^8

max2010maxim [7]

NOTE: The given question is incomplete.

<u>The complete question is given below.</u>

A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.

Solution:

Speed of yellow light in water (v) = 2.00 x 10⁸ m/s

Refractive Index of water with respect to air (μ) = 4/3

Refractive Index = Speed of yellow light in air / Speed of yellow light in water

Or, The speed of yellow light in air = Refractive Index × Speed of yellow light in water

or, = (4/3) × 2.00 x 10⁸ m/s

or, = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s

Hence, the required speed of yellow light in the air will be 3.0 × 10⁸ m/s.

7 0

3 years ago

A low-pass first-order instrument has a time constant of 20 ms. find the frequency, in hertz, of the input at which the output w

Vladimir [108]

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8 0

3 years ago