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zhenek [66]
3 years ago
5

Which is an example of the force of attraction between two objects that have mass?

Physics
1 answer:
Mariulka [41]3 years ago
4 0

Answer:

Gravity is an example of the force of attraction between two objects that have mass.

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yellow cab taxi charges a $1.75 flat rate in addition to $0.65 per mile. kim has no more than $10 to spend on a ride. how many m
Nat2105 [25]

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you could go 12 miles paying $7.80 and $1.75

So in total being $9.55

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Extrusive rocks forms beneath earth's surface true or false
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The answer is false your welcome
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The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for
abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta    - \omega_{0}^{r} \\\\\to \omega^{r} = 2  (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}

5 0
3 years ago
Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
10m= (5.0) + (.5)(9.8)(5.0)
Norma-Jean [14]
M=2.45 because you multiply out the equation on the right and divide by 10
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3 years ago
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