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xz_007 [3.2K]
3 years ago
11

A certain race track is four and a half kilometers long. The winning driver in the race drove his car around the track TWICE in

three minutes. What was his average speed over those two laps? Please answer in kilometers per hour.
HELP!!!!!!!!! BRAINLIEST!!!!!!!!!!
Physics
1 answer:
Vika [28.1K]3 years ago
4 0

he drove 3 kilometers every minute

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JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back
zzz [600]
(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second 

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>
3 0
3 years ago
Two automobiles of equal mass approach an intersection. One vehicle is traveling with velocity 14.0 m/s toward the east and the
Sergeeva-Olga [200]

Answer:

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Explanation:

4 0
3 years ago
The latitude of the city of Arlington is about 32.7357°. Calculate the following: (a) The angular speed of the Earth. (b) The li
julia-pushkina [17]

Answer:

Explanation:

a ) The earth rotates by 2π radian in 24 x 60 x 60 s

so angular speed ( w )  = 2π / (24 x 60 x 60)  = 7.268 x  10⁻⁵ rad / s

b ) Linear speed of city of Arlington ( v )  = w r = w R Cosλ where R is radius of the earth and λ is latitude .

v = 7.268 x 10⁻⁵ x  6.371 x 10⁶ cos 32.7357

389.5 m /s

acceleration = w² r = w² R Cos 32.7357

= (7.268 x 10⁻⁵ )² x 6.371 x 10⁶ x cos 32.7357

=283.08 x 10⁻⁴ m/s²

c) velocity ratio =

w r /w R =

R cos 32.73/ R  

= Cos 32.73

= 0.84 .

6 0
3 years ago
The surface of the Sun has a temperature of about 5 800 K. If the radius of the Sun is 7 × 108 m, determine the power output of
Nimfa-mama [501]

Answer:

a. 3.95\times10^{26}W

Explanation:

T = temperature of the surface of sun = 5800 K

r = Radius of the Sun = 7 x 10⁸ m

A = Surface area of the Sun

Surface area of the sun is given as

A = 4\pi r^{2} \\A = 4(3.14) (7\times10^{8})^{2}\\A = 6.2\times10^{18} m^{2}

e = Emissivity = 1

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the sun is given as

P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (1) (6.2\times10^{18}) (5800)^{4}\\P = 3.95\times10^{26} W

4 0
3 years ago
I drop a meter stick , and catch it as it falls. If it fell exactly 22 cm, how much did it take me to catch the meter stick (wha
statuscvo [17]
D=Vot+1/2at^2

In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2

acceleration=acceleration due to gravity=-9.8m/s^2

It falls - 22cm or -0.22m

We have - 0.22=1/2(-9.8)t^2

t^2=(-0.44)/(-9.8)

t=sqrt[0.44/9.8]
3 0
3 years ago
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