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3 years ago

Does warm water hold more salt then cold water?

Physics

2 answers:

maksim [4K]3 years ago

8 0

yes. ♛┈⛧┈┈•༶༶•┈┈⛧┈♛♛┈⛧┈┈•༶༶•┈┈⛧┈♛

Vinil7 [7]3 years ago

3 0

Answer: Yes

Explanation: Because the water molecules expand when exposed the the warmth of the sun witch means that warm water can carry more salt.

Hope this helps!

Have a good rest of your day!

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A hurricanes energy comes from?

Shalnov [3]

Answer:

a or b

Explanation:

6 0

2 years ago

Read 2 more answers

A truck is traveling east at 80 km/h. At an intersection 32 km ahead, a car is traveling north at 50 km/h. How long after this m

nirvana33 [79]

The time elapsed when the vehicles are closest to each other is 20 min.

The given parameters:

Speed of the truck, u = 80 km/h
Distance, d = 32 km
Speed of the car, v = 50 km/h

<h3>Principles of relative speed</h3>

The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.

$(V_r) t = d\\\\V_r^2 = 50^2 + 80^2\\\\V_r =\sqrt{50^2 + 80^2} \\\\V_r = 94.34 \ km/h$

$94.34 t = 32\\\\t = \frac{32}{94.34} \\\\t = 0.34 \ hr\\\\t \approx 20 \min$

Thus, the time elapsed when the vehicles are closest to each other is 20 min.

Learn more about relative velocity here: brainly.com/question/24430414

3 0

2 years ago

Describe a situation that includes no less than four charges of any magnitude, but they combine so that another location, p, has

Degger [83]

Answer:

Four charges of equal magnitude sitting at the vertices of a square

Explanation:

We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.

Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.

We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.

6 0

3 years ago

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of t

Maslowich

Answer:

a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s

b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s

Explanation:

a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)

It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.

Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).

Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s

b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.

The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,

0.5T = 2.76 s

T = 2 × 2.76 = 5.52 s

Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as

y = A cos wt (provided that there's no phase difference, that is, Φ = 0)

A = amplitude = 0.18 m

w = (2π/5.52) = 1.138 rad/s

When y = 0.09 m, the time = t₁₂ = ?

0.09 = 0.18 Cos 1.138t₁ (angles in radians)

Cos 1.138t₁ = 0.5

1.138t₁ = arccos (0.5) = (π/3)

t₁ = π/(3×1.138) = 0.92 s

When y = -0.09 m, the time = t₂ = ?

-0.09 = 0.18 Cos 1.138t₂ (angles in radians)

Cos 1.138t₂ = -0.5

1.138t₂ = arccos (-0.5) = (2π/3)

t₂ = 2π/(3×1.138) = 1.84 s

Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s

Hope this Helps!!!

3 0

3 years ago

A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of

Fudgin [204]

Answer:

Part a)

$v_f = v_x = 32.77 m/s$

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

$v_x = 40 cos35 = 32.77 m/s$

$v_y = 40 sin35 = 22.94 m/s$

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

$v_f = 32.77 m/s$