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kobusy [5.1K]
3 years ago
13

Calculate the speed for a car that went a distance of 254 km in 4 hours

Physics
1 answer:
quester [9]3 years ago
5 0

Answer:

speed = 63.5 km/hr

Explanation:

let me know if that works

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The cones present on the retina are the cells that help you see the colors.....there are mainly three types....red blue and green, cones but they mix to generate different colours !
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a racing car traveling initially at 8.0 m/s accelerates uniformly at 10.0 m/s^2 for 5 seconds. How far does it travel in this ti
Pepsi [2]

The car travels a distance of

(8.0 m/s) (5 s) + 1/2 (10.0 m/s²) (5 s)² = 165 m

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How much larger is the diameter of Jupiter compared to the diameter of Saturn?
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A point source of light is 1.24 m below the surface of a pool. What is the diameter of the circle of light that a person above t
Ksivusya [100]

Answer:

D = 2.828 m

Explanation:

given,

distance of source of light = 1.24 m below surface of pool

refractive index of the water  = n₁ = 1.33

refractive index of air = n₂ = 1

refraction angle be = 90°

let C be the critical angle

Radius = d tan C

d is the depth of the source

Using Snell's law

n₁ sin C = n₂ sin R

1.33 x sin C = 1 x  sin 90°

sin C = \dfrac{1}{1.33}

C = sin^{-1}(0.752)

C = 48.75°

hence,

R = 1.24 x tan 48.75°

R = 1.414 m

Diameter = 2 x R

D = 2 x 1.414

D = 2.828 m

5 0
3 years ago
Early in 1981 the Francis Bitter National Magnet Laboratory at M.I.T. commenced operation of a 3.3 cm diameter cylindrical magne
Igoryamba

Answer:

The maximum electric field E_{max}= 0.132V/m

Explanation:

From the question we are told that

       The diameter is  d = 3.3 cm = \frac{3.3}{100} = 0.033m

         The magnetic field  of the cylinder is B = 30 T

           The frequency is  f = 15Hz

            The radial distance is  d_r = 1.4cm = \frac{1.4}{100}  = 0.014m

This magnetic field can be represented mathematically as

        B(t) = B_i + B_1sin (wt + \o_i)

 The initial magnetic field is the average between the variation of the magnetic field which is represented as

          B_i = \frac{30 + 29.6}{2}

               =  29.8T

Then B_1 is the amplitude of the  resultant  field is mathematically evaluated as

                 B_1 = \frac{30.0 - 29.6}{2}

                       = 0.200T

The electric field induced can be represented mathematically as

        E = \frac{1}{2}  [\frac{dB }{dt}  ]d_r

            = \frac{d_r}{2} \frac{d}{dt} (B_i + B_1 sin (wt + \o_o))

            = \frac{1}{2} (B wr cos (wt + \o_o))

At maximum electric field  cos (wt + \o_o)  = 1

        E_{max} = \frac{1}{2} B_1 wd_r

         E_{max} = \frac{1}{2} B_1 2 \pi f d_r

                  = \frac{1}{2} (0.200) (2 \pi (15 ))(0.014)

                 E_{max}= 0.132V/m

                   

           

6 0
3 years ago
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