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Nookie1986 [14]
3 years ago
12

Samuel adds a teaspoon of salt to a glass of water. He notices that the salt disappears. Samuel takes a dip to discover that the

water tastes salty. What kind of change has occurred
Physics
1 answer:
Tpy6a [65]3 years ago
6 0
It is a chemical change because the salt is dissolving.
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Match the following vocabulary words with their definitions:
Ronch [10]

Correct matching:


1 acceleration --> rate of change in velocity, which is the change in velocity divided by the change in time

2. speed --> the rate at which an object changes position when traveling in a certain direction

4. gravity --> force of attraction between all masses in the universe

5. Inertia --> an object´s resistance to a change in motion

3. friction --> force of resistance acting between objects in contact and tending to dampen their motion

6. velocity --> the rate at which an object changes position


6 0
3 years ago
What distance will a car cover while uniformly accelerating from 12m/s to 26m/in 14 seconds​
VikaD [51]

Answer:

Distance = 266m

Explanation:

6 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
How much voltage is in the primary coil if there are 3200 windings in the
Lesechka [4]

Answer:

Voltage in primary coil is 3.91 V

Explanation:

For transformer we know that the working principle is given as

\frac{V_1}{V_2} = \frac{N_1}{N_2}

here we know that

V_1 [tex] = voltage in primary coil[tex]V_2 = 25 V

N_1 = 500

N_2 = 3200

Now we have

\frac{V_1}{25} = \frac{500}{3200}

V_1 = 3.91 V

8 0
3 years ago
Consider two identical insulated metal spheres, A and B. Sphere A initially has a charge of -6.0 units and sphere B initially ha
Oksi-84 [34.3K]

Answer:

<em>-2 units of charge</em>

Explanation:

charge on A = Qa = -6 units

charge on B = Qb = 2 units

if the spheres are brought in contact with each other, the resultant charge will be evenly distributed on the spheres when they are finally separated.

charge on each sphere will be = \frac{Qa + Qb}{2}

charge on each sphere =  \frac{-6 + 2}{2} = \frac{-4}{2} = <em>-2 units of charge</em>

8 0
3 years ago
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