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3 years ago

Which best describes why liquids needs a container when a solid does not

Physics

2 answers:

Vika [28.1K]3 years ago

6 0

Answer:

The answer to this question is B: Solids have a definite shape and a liquid takes the shape of its container

Explanation:

I did the quiz

Vladimir [108]3 years ago

4 0

Answer:because liquids does not stick togather so it spreads out and solids don't

Explanation:

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

A boat of mass 225 kg drifts along a river at a speed of 21 m/s to the west. what impulse is required to decrease the speed of t

zzz [600]

The impulse required to decrease the speed of the boat is equal to the variation of momentum of the boat:
$J=\Delta p=m \Delta v$
where
m=225 kg is the mass of the boat
$\Delta v=v_f-v_i=15 m/s-21 m/s=-6 m/s$ is the variation of velocity of the boat
By substituting the numbers into the first equation, we find the impulse:
$J=m\Delta v=(225 kg)(-6 m/s)=-1350 N s$
and the negative sign means the direction of the impulse is against the direction of motion of the boat.

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The buoyant force on any object acts in the direction opposite to the force of gravity. (A)

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