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Contact [7]
2 years ago
14

Speed of a train is 5m/s, find its value in km/hr.​

Physics
2 answers:
antiseptic1488 [7]2 years ago
4 0

Answer: 18km/hr

0.005km/s

1 sec --> 0.005

3600sec --> 0.005*3600 is 18

So, 18km/hr

Free_Kalibri [48]2 years ago
3 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Here's the solution ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:5 \: m \: s {}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:\dfrac{5}{1000} km \:  {s}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:\dfrac{5}{1000} \times 3600 \:  km \:  {h}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{5}{5}  \times 18 \: km \: h {}^{ - 1}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: 18 \: km \: h {}^{ - 1}

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How much heat must be added to make a 5g substance with a specific heat of 2 J/gC that has its temperature go up 10 degrees? Q =
zlopas [31]

Answer:

100 Joule

Explanation:

Amount of heat in agiven body is given by Q = m•C•ΔT

where m is the mass of the body

c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.

ΔT is the change in the temperature of body

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coming back to problem

m = 5g

C = 2J/gC

since, it is given that temperature of body increases by 10 degrees, thus

ΔT = 10 degrees

Using the formula for heat as given

Q = m•C•ΔT

Q = 5* 2 * 10  Joule= 100 Joule

Thus, 100 joule heat must be added to  a  5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.

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3 years ago
A planet moves forward because of momentum <br><br> True or False??
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true

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3 years ago
Read 2 more answers
g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0
3 years ago
Read 2 more answers
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