Which term measures how much salt will dissolve into 100 g of water?

Physics

1 answer:

nataly862011 [7]3 years ago

7 0

Answer:

The answer is Solubility

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Where are variables represented in a graph?

m_a_m_a [10]

Answer:variable or dependent values are represented on y axis of graph

Explanation:

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3 years ago

A 1.0 104 kg spacecraft is traveling through space with a speed of 1200 m/s relative to Earth. A thruster fires for 2.0 min, exe

aniked [119]

We are given information:
$m=1.0* 10^{4} kg \\ v=1200m/s \\ t=2min=120s \\ F = 25kN = 25000N$

If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2

Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s

Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s

Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s

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3 years ago

Green plants give us oxygen to breathe using photosynthesis. Here is the unbalanced chemical reaction that occurs:

bekas [8.4K]

Books are always useful . Give me a "thank you" if i've been helpful

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A spherical conducting shell has charge Q. A point charge q is placed at the center of the cavity. The charge on the inner surfa

FinnZ [79.3K]

Answer:d

Explanation:

From Gauss law Electric field inside a surface is directly proportional to the charge enclosed in it.

Electric field inside a spherical shell is zero and hence there is no charge inside the spherical shell because q charge induces a -q charge on inside surface of spherical shell.

and to counter it there is q charge on the surface. So total charge outside the surface is Q+q

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3 years ago

If an athlete leaps vertically at 4.0m/s, what maximum height does he reach?

joja [24]

Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration: $g=-9.81~m/s^2$ , and where the negative sign means it points downwards, against the direction of the motion.

Therefore, we can use the following formula to solve the problem:
$v_f^2 = v_i^2 + 2gS$
where $v_i=4~m/s$ is the initial vertical velocity of the athlete, $v_f=0$ is the vertical velocity of the athlete at the maximum height (and $v_f=0~m/s$ at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
$S= \frac{v_f^2-v_i^2}{2g}=0.82~m$

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3 years ago