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2 years ago

PLEASE HELP!!

Physics

1 answer:

leva [86]2 years ago

5 0

Answer:

C 1.67 × 10-9 N

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1 ,m2 are the masses of the two objects

r is their separation

In this problem, we have:

$m_1 = m_2 = 15.0 kg$

r = 3.00 m

Substituting into the equation, we find

$F=(6.67\cdot 10^{-11}) \frac{(15)(15)}{3.00^2}=1.67\cdot 10^{-9} N$

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I hope it helps you!

4 0

3 years ago

physics A river flows at a speed vr = 5.37 km/hr with respect to the shoreline. A boat needs to go perpendicular to the shorelin

vova2212 [387]

Answer: Vb is the vector (-5.37m/s, 8.59 m/s), with a module 10.13m/s

then the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9

Explanation:

We can use the notation (x, y) where the river flows in the x-axis and the pier is on the y-axis.

We have Vr = (5.37m/s, 0m/s)

Now, if the boat wants to move only along the y-axis (perpendicularly to the shore).

The velocity of the boat Vb will be:

Vb = (-c*sin(32). c*cos(32))

Then we should have that:

5.37 m/s - c*sin(32) = 0

c = (5.37/sin(32))m/s = 10.13 m/s

the velocity in the y-axis is:

10.13m/s*cos(32) = 8.59 m/s

So Vb = (-5.37m/s, 8.59 m/s)

the ratio Vb/Vr = 10.13m/s/5.37m/s = 1.9 where i used Vb as the module of the boat's velocity.

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3 years ago