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wariber [46]
3 years ago
14

1. A body has a velocity of 72 km/hr. Find its value in m/s.

Physics
1 answer:
CaHeK987 [17]3 years ago
8 0

72 Km/hr

= 72000 m/ 60×60 s

= 72000 m/ 3600 s

= 20 m/s

Answer is 20 m/s.

Hope it helps! Please do comment

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<h3>\huge\underline\bold\blue{ƛƝƧƜЄƦ}</h3><h3>Given</h3>

\blue\star v = 20m\s

\blue\star a = 3m\s^2

\blue\star t = 4sec

Firstly we have to find u

\star a = \dfrac{v - u}{t}

\star 3m\s =\dfrac{20 - u}{4}

\star12m\s = 20 - u

\star20 - u = 12m\s

\star- u = -8

\star u = 8

Now we can easily find distance by using second equation of motion

\red\stars = ut + 1\2 at^2

\red\stars = 8(4) + 1\2(3)(16)

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3 years ago
A plane moves at a certain velocity and then it accelerates at the rate of 52m/s for a distance of 2300m to reach a velocity of
Illusion [34]
52m/s is the answer because before it was gonna accelerate it was 52 m/s
6 0
3 years ago
Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black
oksian1 [2.3K]

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.

Heat flow is obtained as follows:

Q = FA\sigma\Delta T^4

Where,

F =View Factor

A = Cross sectional Area

\sigma = Stefan-Boltzmann constant

T= Temperature

Our values are given as

D = 0.6m

L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K

The view factor between two coaxial parallel disks would be

\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33

\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75

Then the view factor between base to top surface of the cylinder becomes F_{12} = 0.26. From the summation rule

F_{13} = 1-0.26

F_{13} = 0.74

Then the net rate of radiation heat transfer from the disks to the environment is calculated as

\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}

\dot{Q_3} = 2\dot{Q_{13}}

\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)

\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)

\dot{Q_3} = 780.76W

Therefore the rate heat radiation is 780.76W

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Which of the following describes resistance?
enot [183]

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How does the height of a ramp affect the distance that a toy car will travel?
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