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nikdorinn [45]
3 years ago
10

Susan goes out to exercise. She runs for one hour at a constant speed and velocity. What is the overall net force acting on Susa

n?
Physics
2 answers:
umka21 [38]3 years ago
7 0
Well since there is no acceleration..the net force is zero
Roman55 [17]3 years ago
3 0
<span>The net force acting on the runner must be zero.</span>
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Genes are organized on matching pairs of _______.
natta225 [31]

Answer:

homologous chromosomes

Explanation:

6 0
3 years ago
If you drop a silver dollar off a building and it hits the ground in 10 seconds, how fast was the coin going just before?
Ksju [112]
V = u + at 
<span>= 0 + (9.81)(10) </span>
<span>= 98.1m/s </span>

<span>Ignoring air resistance.</span>
3 0
3 years ago
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
What did the results of photoelectric-effect experiment establish?
Bond [772]

Answer:

Thanks For Points Mark Me Brainliest Please

Explanation:

Electrons are emitted as soon as low-intensity, high-frequency light hits a metal surface.

A. Is It

7 0
3 years ago
Read 2 more answers
Develop an equation (with a proportionality constant) that describes the relationship between the gravitational force (fgrav), t
ivolga24 [154]

According to newton's law of gravitation, the gravitational force(F)  is directly proportional to the product mass of the moon(Mm) and the mass of the planet (Mp) and it is inversely proportional to the square of the separation between them.

Fg ∝ (Mp)(Mm) →(1)

Fg ∝ 1/d²→(2)

Combining equation (1) and (2),

Fg ∝ (Mp)(Mm)/d²

Fg =  G(Mp)(Mm)/d²

This is an equation that describes the relation between mass of moon (Mm) and mass of planet (Mp) and separation(d) between them.

To support the claim in favuor of this equation we use this equation to obtain the value of acceleration due to gravity on earth.

Let m be the mass of an object on earth then Fg between earth (Mp) and mass of an object is obtained by:

Fg =  G(Mp)(m)/R², where R= Radius of earth

This force is equal to the weight of an object i.e.,

g= G(Mp)/R²

Putting the values of G, Mp and R , we get, g=9.81 m/s²

which is the value we obtained on earth for acceleration due to gravity.

To know more about gravitational constant, visit:

brainly.com/question/13959861

#SPJ4

5 0
1 year ago
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