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Verizon [17]
3 years ago
8

In the design of a new building, a doorway is 2.45 ft above the ground. A ramp for the disabled, at an angle of 4.5 ° with the g

round, is to be built up to the doorway. How long will the ramp be?

Physics
1 answer:
kkurt [141]3 years ago
3 0

Answer:

Length of the ramp, l = 31.22 feet

Explanation:

It is given that,

A doorway is 2.45 ft above the ground, AB = 2.45 ft

Angle between the ground and the ramp is, \theta=4.5^{\circ}

Applying trigonometry,

sin\theta=\dfrac{AB}{AC}

AC=\dfrac{AB}{sin\theta}

AC=\dfrac{2.45}{sin(4.5)}

AC = 31.22 ft

So, the length of the ramp is 31.22 feet. Hence, this is the required solution.

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The human eye can readily detect wavelengths from about 400 nm to 700 nm. part a if white light illuminates a diffraction gratin
Lera25 [3.4K]

We are given that the wavelength ʎ is from 400 nm to 700 nm. The formula for this is:

d sin a =m * ʎ

where,

d = slit separation = 1 mm / 750 lines = 1/750

a = angle

m = 1

ʎ = 400 nm to 700 nm = 0.0004 mm to 0.0007 mm

 

Rewriting the formula in terms of angle a:

a = sin^-1 (m ʎ / d)

 

when ʎ = 0.0004 mm

a = sin^-1 (0.0004 / (1/750))

a = 17.46°

 

when ʎ = 0.0007 mm

a = sin^-1 (0.0007 / (1/750))

a = 31.67°

 

Hence the range of angles is from 17.46° to 31.67<span>°.</span>

3 0
3 years ago
An object weighs 15N in air and 13N when completely
anastassius [24]

Answer:

The volume of water displaced = 0.204 m³

Explanation:

<em>From Archimedes' principle,</em>

<em>Upthrust = lost in weight = weight of water displaced.</em>

<em>U = W₁ - W₂..................... Equation 1</em>

<em>Where U = upthrust, W₁ = weight in air, W₂ = weight when completely submerged in water.</em>

<em>Given: W₁ = 15 N, W₂ = 13 N</em>

<em>Substituting these values into equation 1</em>

<em>U = 15 - 13 </em>

<em>U = 2 N</em>

<em>But,</em>

<em>Weight of water displaced = mass of water displaced. × acceleration due to gravity.</em>

<em>Mass of water displaced = weight of water displaced/ acceleration due to gravity.................... Equation 2</em>

Where: acceleration due to gravity = 9.8 m/s², weight of water displaced = 2 N

Substituting these values into equation 2

Mass of water displaced = 2/9.8

Mass of water displaced = 0.204 kg.

Also,

Volume of water displaced = mass of water displaced/ Density of water.............. Equation 3

Where Density of water = 1.0 kg/m³, mass of water displaced = 0.204 kg

Substituting these values into equation 3 above,

Volume of water displaced = 0.204/1

volume of water displaced = 0.204 m³

Therefore the volume of water displaced = 0.204 m³

<em />

8 0
3 years ago
The net torce on an object moving with constant speed in circular motion is in which direction?
aleksley [76]

The correct answer is C) towards the center of the circle.

Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.

For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.

6 0
3 years ago
A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski
Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0
3 years ago
An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the soleno
GaryK [48]

Answer:

The maximum change in  flux is \Delta \o = 0.1404 \ Wb

The average  induced emf     \epsilon =0.11232 V

Explanation:

   From the question we are told that

             The speed of the technician is v = 0.80 m/s

              The distance from the scanner is d = 1.0m

              The  initial magnetic field is  B_i = 0T

               The final magnetic field is B_f = 6.0T

                 The diameter of the loop is  D = 19cm = \frac{19}{100} = 0.19 m

The area of the loop is mathematically represented as

        A  =  \pi [\frac{D}{2} ]^2

             = 3.142 \frac{0.19}{2}

             = 0.02834 m^2

At maximum the change in magnetic field is mathematically represented as

            \Delta \o = (B_f - B_i)A

  =>      \Delta  \o = (6 -0)(0.02834)

                  \Delta \o = 0.1404 \ Wb

The  average induced emf is mathematically represented as

           \epsilon =  \Delta \o v

              = 0.1404 * 0.80

             \epsilon =0.11232 V

7 0
3 years ago
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