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taurus [48]
3 years ago
7

A chemistry student mixes 5 grams of sodium chloride (NaCl) in 100 ml of water and stirs until all of the salt is dissolved. Onc

e dissolved, the student measures the electrical conductivity of the solution. An additional 5 grams of NaCl is then dissolved in the solution. Which statement MOST accurately describes how the additional five grams of salt will affect the electrical conductivity of the solution? A) The conductivity of the solution will increase because more ions are being added to the solution. B) The conductivity of the solution will decrease because salt is an insulator and more is being added. C) The conductivity of the solution will decrease because the concentration has changed--the solution is more dense. D) The conductivity of the solution will increase because additional electrical potential is being added to the solution.
Physics
1 answer:
vichka [17]3 years ago
4 0
The answer would be a
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An atomic nucleus is composed of
musickatia [10]

Answer:

84 protons and 128 neutron

8 0
3 years ago
Please help it’s easy will give brainlist !! Thank youu worth a lot be serious pls
Genrish500 [490]

Answer:

Part 1

Stationary

Part 2

20 Newtons

Part 3

Force

Part 4

4.0 m/s²

Part 5

Normal

Part 6

Cart m

Part 7

The gravitational force is less than magnetic force

Explanation:

Part 1

The position time graph of the object is an horizontal straight line passing across the top of the position 3 boxes vertically up from the origin

As the time increases by the units of number boxes to the left, the position of the object does not change and remains at the 3 boxes up above the origin, therefore, the object is stationary

Part 2

By Newton's third law of motion, the action action obtained from a force is equal to the reaction given to the force, therefore, we have;

The force exerted by the student on the scale = The force exerted by the scale on the student = 20 N

Part 3

A force is a the directional push on an object or pull from the object as a form of interaction with another object which tends to alter or maintain the motion of the object

Part 4

The given parameters are'

The mass of block A = 1.0 kg

The mass of block B = 2.0 kg

Both blocks, "A" and "B" are initially at rest

The applied horizontal force, F = 12-N

The nature of the surface over which the blocks move = Smooth surface

Force, F = Mass, m × Acceleration, a

F = m × a

The blocks two blocks experience a common acceleration, a

The combined mass of the two blocks, m = 1.0 kg + 2.0 kg = 3.0 kg

m = 3.0 kg

Therefore, a = F/m = 12-N/(3.0 kg) = 4 m/s²

Part 5

A normal force is a force acting perpendicularly to a surface that supports the weight of an object

Part 6

The given parameters are;

M = 2 kg

The mass of the left cart attached to the spring = M = 2 kg

The mass of the right cart attached to the spring = 2·M = 2 × 2 kg = 4 kg

Therefore, given that the force exerted by one cart on the other after the spring is removed, we have;

Force, F = M × a₁ = 2·M × a₂

Where;

a₁ = The average acceleration of the cart with mass, M

a₂ = The average acceleration of the cart with mass, 2·M

M × a₁ = 2·M × a₂

∴ a₁ = 2·a₂

The acceleration, a₁, of car M = 2 × The acceleration, a₂, of car 2·M

The acceleration of cart M is two times the acceleration of cart 2·M

Cart M will experience a greater average acceleration

Part 7

For the top magnet of two magnets placed two magnets arranged so that one is on top the other, to be held in the air with a space between the two magnets, the gravitational force pulling the top magnet down, is less than the repelling magnetic force between the two magnets.

7 0
3 years ago
The y-component of a projectile’s velocity is 12.1 m/s. When the projectile once again passes by the height from which it was la
Nat2105 [25]
It's 12.1 m/s, assuming that's the launch velocity that's given.
For projectile motion, velocity's y-component is parabolic/quadratic. It's x-component is constant, so you don't need to know it. 
6 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0
3 years ago
A block of mass m=9.0 kg and speed V and is behind a block of mass M= 27 kg and speed of .50 m/s. The surface is frictionless, a
sammy [17]

Answer:

2.06 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5

Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

V=2.055555555555555555555555555555555555555 m/s

Rounded off, v is approximately 2.06 m/s

5 0
3 years ago
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