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3 years ago

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When you drop a 0.42 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s 2 toward the earth’s surface. According

to Newton’s third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth 5.98 × 1024 kg, what is the magnitude of the earth’s acceleration toward the apple? Answer in units of m/s 2 .

2 answers:

Murljashka [212]3 years ago

4 0

Answer:

6.9×10⁻²⁵ m/s²

Explanation:

m = Mass

a = Acceleration

F = ma

Force exerted on apple

$F=0.42\times 9.8\\\Rightarrow F=4.116\ N$

From Newton's third law the opposite force would be the same but the mass is different which means the acceleration would be different

$a=\frac{F}{M}\\\Rightarrow a=\frac{4.116}{5.98\times 10^{24}}\\\Rightarrow a=6.9\times 10^{-25}\ m/s^2$

The magnitude of the earth’s acceleration toward the apple is 6.9×10⁻²⁵ m/s²

zepelin [54]3 years ago

3 0

Answer:

a = 6.97 × 10⁻²⁵ m/s²

Explanation:

given,

mass = 0.42 kg

g = 9.8 m/s²

mass of the earth = 5.98 × 10²⁴ kg

magnitude of earth acceleration = ?

F = m g

F = 0.42 × 9.8

F = 4.17 N

according to newtons third law Force exerted by the earth will be equal to force exerted by apple

m a = 4.17 N

$a = \dfrac{4.17}{5.98 \times 10^{24}}$

a = 0.697 × 10⁻²⁴ m/s²

a = 6.97 × 10⁻²⁵ m/s²

Hence, acceleration of earth toward apple is equal to a = 6.97 × 10⁻²⁵ m/s²

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a.

$\displaystyle a(0 )=8.133\ m/s^2$

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Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

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Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

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$\displaystyle a(0)=8.133547\ e^o$

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For t=2

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$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

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The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

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