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2 years ago

Crushing, stirring, heating are all methods to increase the rate of ______________ a solute in a solvent.

Physics

1 answer:

Fofino [41]2 years ago

4 0

Answer: mixing

Explanation:

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How can you keep sound out of a room

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3 years ago

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

$V_{2}=11369.46 m/s$

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2 years ago

A boy is going down a slide. As he reaches the bottom, friction causes him to slow down and stop. Which one of Newton's laws is

lara31 [8.8K]

I think it’s the second law

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3 years ago

What happens if two plates move away from each other?

Sloan [31]

Answer:

They will sometimes crash into other plates in the process and will rub while they are moving creating earthquakes

Explanation:

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2 years ago

Read 2 more answers

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

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3 years ago