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3 years ago

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(a) calculate earth's mass given the acceleration due to gravity at the north pole is 9.830 m/s2 and the radius of the earth is

6371 km from center to pole. (b) compare this with the accepted value of 5.979×1024 kg .

1 answer:

kakasveta [241]3 years ago

5 0

(a) The value of the gravitational acceleration is given by:

$g=\frac{GM}{r^2}$

where

$G=6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

M is the Earth's mass

$r=6371 km=6.371 \cdot 10^6 m$ is the Earth's radius at the pole

If we use the value for g given by the problem, $g=9.830 m/s^2$ , and we rearrange the equation above, we find the value of the Earth's mass:

$M=\frac{gr^2}{G}=\frac{(9.830 m/s^2)(6.371 \cdot 10^6 m)^2}{6.67 \cdot 10^{-11} m^3 kg^{-1}s^{-2}}=5.982 \cdot 10^{24} kg$

(b) The value we found for the Earth's mass is $5.982 \cdot 10^{24} kg$ , and we see that this value is slightly larger than the accepted value of $5.979 \cdot 10^{24} kg$ .

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The speed of the wave with the given frequency and wavelength is 1.2m/s.

Given the data in the question;

Wavelength of the wave; $\lambda = 0.4m$

Frequency; $f = 3Hz = 3s^{-1}$

Speed of the wave; $v = \ ?$

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Where v is velocity/speed and f is frequency.

Now, we can easily get the speed of the wave by substituting our given values into the expression above.

$\lambda = \frac{v}{f} \\\\0.4m = \frac{v}{3s^{-1}}\\ \\v = 0.4m * 3s^{-1}\\\\v = 1.2ms^{-1}\\\\v = 1.2m/s$

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A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

a)

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

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b)

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

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