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ira [324]
3 years ago
11

A car is traveling at 20 m/s. It takes the car 120s to go from one intersection to a second. What was the displacement of the ca

r between the two intersections?
Physics
1 answer:
Bad White [126]3 years ago
7 0

We don't know if the road is perfectly straight between the intersections,  or if the road bends, curves, or turns between them.  So we don't have enough information to calculate the displacement between them.

But we <em>can</em> calculate the <u>distance</u> the car traveled between them.

Distance = (speed) (time)

Distance = (20 m/s) (120 s)

<em>Distance = 2,400 meters</em>

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Define bond length using your own words​
Brut [27]

Answer:

the free encyclopedia. In molecular geometry, bond length or bond distance is defined as the average distance between nuclei of two bonded atoms in a molecule. It is a transferable property of a bond between atoms of fixed types, relatively independent of the rest of the molecule.

Explanation:

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3 years ago
Which is a likely use for a base? a. as a vitamin in your food b. etching metals for printing c. making foods taste sour d. maki
natta225 [31]
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5 0
3 years ago
A satellite originally moves in a circular orbit of radius R around the Earth. Suppose it is moved into a circular orbit of radi
sergiy2304 [10]

 satellite originally moves in a circular orbit of radius R around the Earth. Suppose it is moved into a circular orbit of radius 4R.

(i) What does the force exerted on the satellite then become?

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3 0
3 years ago
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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
A 200g block on a 50cm long string swings in a circle, it's frictionless and 75rpm. What is its speed and tension on string
krek1111 [17]
Angular velocity = (75x2pie)/60
                          =2.5pie ras^-1 
linear velocity(or speed) at end of string, v = radius x angular velocity
                                                           v= 0.5 x 2.5pie
                                                           v=3.93 ms^-1

tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
                                                                                      F= (0.2 x 3.93^2)/0.5
                                                                                      F=6.18 N
(sorry if wrong)
3 0
3 years ago
Read 2 more answers
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