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2 years ago

Which process is an example of a physical change?

Physics

1 answer:

Anit [1.1K]2 years ago

8 0

Answer:

Option A

Carrots are cut into small pieces and mixed into a salad

Explanation:

When physical changes occur, the actual composition remain the same but the molecules are re-arranged. Therefore, when carrots are cut into smaller pieces and mixed into salad, there will be no chemical reaction hence the actual composition will remain the same despite being cut and molecules in it re-arranged. Considering the other options, new substances are formed hence they are deemed as chemical changes. Therefore, option A is correct.

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The A string of a violin is a little too tightly stretched. Beats at 4.00 per second are heard when the string is sounded togeth

Basile [38]

Answer:

$T=2.5*10^{-3}s$

Explanation:

From the question we are told that:

Beat frequency $F_b=4$

Frequency $F=400Hz$

Generally the equation for Frequency of the violin is mathematically given by

$f_v=F_b+F$

$f_v=4+400Hz$

$f_v=404Hz$

Therefore the period of the violin string oscillations is

$T=\frac{1}{f_v}$

$T=\frac{1}{404}$

$T=2.5*10^{-3}s$

3 0

2 years ago

Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an

bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

$\frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}$ ......... (1)

Hence, total charge will be as follows.

$q_{1} + q_{2}$ = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

and, $q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}$

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

= $\frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}$

= 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

5 0

3 years ago

A uniform electric field is directed parallel to the +y axis. If a positive test charge begins at the origin and moves upward al

lukranit [14]

Answer: option 1 : the electric potential will decrease with an increase in y

Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below

V = kq/y

Where k = 1/4πε0

Where V = electric potential,

k = electric constant = 9×10^9,

y = distance of potential relative to a reference point, ε0 = permittivity of free space

q = magnitude of electronic charge = 1.609×10^-19 c

From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.

We have that

V = k/y

We see the potential(V) is inversely proportional to distance (y).

This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.

This fact makes option 1 the correct answer

3 0

3 years ago

I’ll give brainliest!! please help and answer correctly! plsss answer quick

Rashid [163]

Answer: The motion of the object will remain the same

Explanation:

6 0

3 years ago

A truck skids for a distance of 25 m with the road pushing on its tires with force of 1500 N as its brakes

MakcuM [25]

Answer:-6800J

Explanation: 8.0m x 850N = 6800

Rewritten as a negative when brake/stop -6800

6 0

2 years ago

Read 2 more answers