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lorasvet [3.4K]
3 years ago
9

A car traveling at a speed of 13 meters per second accelerates uniformly to a speed of 25 meters per second in 5.0 seconds.

Physics
1 answer:
Vlad1618 [11]3 years ago
4 0
Since car is accelerating uniformly that means we it is accelerating at constant rate if you can say that way. To find its acceleration all we have to do is find difference between v1=13 m/s and v2 = 25m/s   and divide it by interval that car is taking to accelerate.

a = (v2-v1)/t = 2,4m/s^2

since car is accelerating at constat rate and truck covers same distance as car needs to accelerate from 13 to 25m that means that we can say that in that period truck travelled at cars average speed.

Vtruck = (v1 + v2)/2 = 19m/s
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When passing from a low index of refraction medium to a high index of refraction medium which way does the refracted ray bend: t
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toward the normal

Explanation:

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Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

Light travels at a slower speed in water as compared to air because there are more number of interfering molecules in the path of the light in case of water as compared to liquid.  

When a light travels from lower denser medium say water to higher denser medium say water, it bends towards the perpendicular (normal) as its speed reduces in that medium.

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Why are latitude and longitude included on maps
Veseljchak [2.6K]

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to locate places on earth

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2 years ago
What is the equation for an inelastic collision
abruzzese [7]
M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.
7 0
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A liquid of density 1200kg/m³ is filled in a beaker upto the depth of 50 cm. calculate the pressure at bottom of the breaker.(g=
sweet-ann [11.9K]
  • Density=1200kg/m^3=p
  • Height=50cm=0.5m=h
  • Acceleration due to Gravity=9.8m/s^2=g

\boxed{\sf Pressure=pgh}

\\ \qquad\quad\sf{:}\dashrightarrow Pressure=1200(0.5)(9.8)

\\ \qquad\quad\sf{:}\dashrightarrow Pressure=600(9.8)

\\ \qquad\quad\sf{:}\dashrightarrow Pressure=5880Pa

6 0
3 years ago
Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2
igomit [66]

Answer

Applying Wein's displacement

\lamda_{max}\ T = 2898 \mu_mK

1) for sun T = 5800 K

      \lambda_{max} = \dfrac{2898}{5800}

      \lambda_{max} = 0.5 \mu_m

2) for tungsten T = 2500 K

      \lambda_{max} = \dfrac{2898}{2500}

      \lambda_{max} = 1.16 \mu_m

3) for heated metal T = 1500 K

      \lambda_{max} = \dfrac{2898}{1500}

      \lambda_{max} = 1.93 \mu_m

4) for human skin T = 305 K

      \lambda_{max} = \dfrac{2898}{305}

      \lambda_{max} = 9.50 \mu_m

5)  for cryogenically cooled metal T = 60 K

      \lambda_{max} = \dfrac{2898}{60}

      \lambda_{max} = 48.3 \mu_m

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ              0.01           0.4               0.7                 100

λT             58           2320            4060             5.8 x 10⁵

F                0             0.125             0.491                1

fractions

for UV = 0.125  

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509  

8 0
3 years ago
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