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ikadub [295]
3 years ago
13

Therefore an answer in U.S. customary units, such as miles per hour would not be accepted as correct. A car traveling with const

ant speed travels 150 kmkm in 7200 ss. What is the speed of the car
Physics
2 answers:
allochka39001 [22]3 years ago
6 0

Answer:

The speed of the car is 20.84 m/s.

Explanation:

Given that,

The distance covered by the car, d = 150 km

Time, t = 7200 s

We need to find the speed of the car. We know that the speed of an object is given by total distance divided by total time taken. Mathematically,

v=\dfrac{d}{t}\\\\v=\dfrac{150\times 10^3\ m}{7200\ s}\\\\v=20.84\ m/s

So, the speed of the car is 20.84 m/s.

pychu [463]3 years ago
3 0

Answer:

calculate it. its not cheating, its like flash cards but better, besides its made for that reason.;)

Explanation:

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Kipish [7]

Answer:

okay

Explanation:

6 0
2 years ago
Read 2 more answers
How much voltage is required to run 0.64 A of current through a 240 resistor? Use V= IR.
SVETLANKA909090 [29]
B. If you press that into a calculator it comes up with 153.6. You then shift the decimal point 2 times forward and you end up getting 1.5 x 10^2 V.
7 0
3 years ago
Read 2 more answers
HELP PLEASE BOYLES LAW
kogti [31]

Answer:

3 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 0.75 atm

Final pressure (P₂) = 0.5 atm

Final volume (V₂) =?

Using the Boyle's law equation, the new volume (i.e final volume) of the Ne gas can be obtained as:

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 0.75 atm

Final pressure (P₂) = 0.5 atm

Final volume (V₂) =?

P₁V₁ = P₂V₂

0.75 × 2 = 0.5 × V₂

1.5 = 0.5 × V₂

Divide both side by 0.5

V₂ = 1.5 / 0.5

V₂ = 3 L

Thus, the new volume of the Ne gas is 3 L

7 0
3 years ago
If measuring experimental results, what can you predict about the work output of a 1200 Watt hair dryer?
Elden [556K]
I am 90% sure it’s A
3 0
3 years ago
Read 2 more answers
Un the way to the moon, the Apollo astro-
kherson [118]

Answer:

Distance =  345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2}  } \\

F_{m} =G*\frac{m_{m}*m_{a}  }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]

When we match these equations the masses cancel out as the universal gravitational constant

G*\frac{m_{e} *m_{a} }{r_{e}^{2}  } = G*\frac{m_{m} *m_{a} }{r_{m}^{2}  }\\\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2}  }

To solve this equation we have to replace the first equation of related with the distances.

\frac{m_{e} }{r_{e}^{2}  } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m}  )^{2}  } = \frac{7.36*10^{22}  }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17}  \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c }  }{2*a}\\  where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) }  }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2}  } \\a=3.33*10^{19} [m/s^2]

6 0
3 years ago
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