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2 years ago

What happens when you drop a sugar cube into a mug of hot tea

Physics

2 answers:

azamat2 years ago

8 0

The sugar cube disolves

Yanka [14]2 years ago

4 0

The answer is the molecules in the sugar move from an area of higher concentration to the lower concentration.

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A man dropped a dime in a wishing well , he heard it 5 sec later ..Find distance traveled if moving at 10 m/s 2

Alenkasestr [34]

Answer:

The distance traveled is 109.58 m

Explanation:

The Speed of sound in air = 344 m/s

Let the time in which the dime dropped by the man reach an impact in the well = t₁

Let the time in which the sound travel from the well to the man = t₂

Then

1/2× 10 × t₁² = 344 × t₂ which gives;

5 × t₁² = 344 × t₂.........................(1)

Also the total time before the man heard the dime = t₁ + t₂ = 5

Therefore;

t₂ = 5 - t₁

Substituting the value of t₂ in equation (1), we have;

5 × t₁² = 344 × (5 - t₁)

5·t₁² + 344·t₁ - 1720 = 0

Using the quadratic formula, we have;

$x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}$

Which gives;

$t_1 = \dfrac{-344\pm \sqrt{344^{2}-4\times 5\times (-1720)}}{2\times 5}$

t₁ = 4.68 s or -73.48 s

Therefore, with the positive value for t₁ = 4.68 s, we have

The distance = 1/2× 10 × 4.68² = 109.58 m

The distance traveled = 109.58 m.

8 0

3 years ago

For an object moving with constant negative acceleration, draw the following:

Tcecarenko [31]

Answer:

Kindly find the graphs attached

Explanation:

For figure 1: There is a steady increase in the position of the object as time increases. This is because despite the negative acceleration (deceleration), the object continues to move and cover more ground as time goes by.

The straight line graph is observed because the acceleration is constant and not varying.

For Figure 2: The graph of velocity vs time will have an inverted nature. This is because since the object is decelerating, it is reducing in its velocity as time goes by (increases). This is also in a straight line since the deceleration is constant.

3 0

2 years ago

What is the relationship between temperature, dew point temperature, and air pressure

olchik [2.2K]

Dew point is the temperature at which the water vapor in the air condenses , then evaporates. The barometric or air pressure is independent from the dew point.

6 0

2 years ago

Which bike rider has the greatest momentum?

Sphinxa [80]

My guesses would be g. or F

5 0

2 years ago

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

Given :

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( $K_{A}$ = 0) to the point B at distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

Required :

(a) We are asked to find the electric potential VB

(b) We want to determine the magnitude and the direction of the electric field E.

Solution

(a) We are given the values for VA, $K_{B}$ and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential. So, equation (1) will be in the form :

0+qVA= $K_{B}$ +qVB (Divide by q)

VA= $K_{B}$ /q + VB (solve for VB)

VB=VA- $K_{B}$ /q .......................................(2)

We get the relation between VB, VA and $K_{B}$ , now we can plug our values for VA, $K_{B}$ and q into equation (2) to get VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

=80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (solve for E)

E= VA-VB/ $l$ ..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

E= VA-VB/ $l$

=-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0

2 years ago