Super Mario and Bowser Jr. are racing around a track when Baby Bowser launches a green shell at Mario, bringing him to rest. Bowser Jr. then passes Mario at his top speed of 30 blocks/h, moving down the track in a straight line. Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track. Mario world measure distance using the units of blocks, with 1 block = 0.47 m.

a) What are Mario and Bowser Jr.'s speeds in m/s?

Assuming both Mario and Bowser Jr. race to the finish in a straight line at their top speeds,

b) How long does it take for Mario to catch Bowser Jr.?

c) How far down the track is Mario from the point at which he reaches his top speed?


Answer 1


(a). Mario's speed in m/s = 5.2 × 10^-3 m/s.

Bowser Jr.'s speeds in m/s = 3.92 × 10^-3 m/s.

(b). 27001.2 seconds(s)..

(c). 141 metre(m).


So, the following data or parameters or information was given in the question above. These informations are going to help us in solving this question or problem;

=>" Bowser Jr. then passes Mario at his top speed = 30 blocks/h.

=> " Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track."

=> "Mario world measure distance using the units of blocks, with 1 block = 0.47 m"

Therefore, the solution is given below;

(1). For the first part, we are to determine or calculate Mario and Bowser Jr.'s speeds in m/s.

Therefore, Mario's speed in m/s = 40 × 0.47) ÷ 3600 = 5.2 × 10^-3 m/s.

Also, Bowser Jr.'s speeds in m/s = ( 30 × 0.47) ÷ 3600 = 3.92 × 10^3 m/s.

(2). So, the next thing to do to determine or calculate how long it took for Mario to catch Bowser Jr.

Thus, the time it took for Mario to catch Bowser Jr. Can be related as below;

[ ( 5.2 × 10^-3 m/s) - (3.92 × 10^-3 m/s) × (time,t taken for Mario to catch Bowser Jr.) = 75 × 0.47.

Therefore, the time it took for Mario to catch Bowser Jr. = 27001.2 seconds.

(3). Now, we calculate How far down the track Mario from the point at which he reaches his top speed.

The distance = 5.2 × 10^-3 m/s × 27001.2m = 141m

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Best regards.

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The energy (E) of a photon is:

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[tex] E = h\frac{c}{\lambda} = 6.62 \cdot 10^{-34} J.s\frac{3.00\cdot 10^{8} m/s}{8.97 \cdot 10^{-2} m} = 2.21 \cdot 10^{-24} J [/tex]

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I hope it helps you!

The energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.

Calculation of energy:

We know that

[tex]E = h\frac{c}{\lambda}[/tex]


h be the Planck's constant = 6.62x10⁻³⁴ J.s

λ be the wavelength of the radiation = 8.97 cm

c be the speed of light = 3.00x10⁸ m/s


Here we need to multiply the answer 10^25 so that the correct answer could come.

[tex]E = 6.62.10^{-34} \frac{3.00.10^{8}}{8.97.10^{-2}}[/tex]

=  2.21x10⁻²⁴ J.

= 22.10 J.

Hence, the energy of one photon is 2.21x10⁻²⁴ J. Multiplied by 10²⁵ is 22.10 J.

learn more about energy here;

Question 4
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hello your question is incomplete attached below is the complete question

answer : Ax = 126.13 N

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attached below is the detailed solution  

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Data obtained from the question include:

Mass of satellite (m) = 220 Kg

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