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Degger [83]
3 years ago
6

Which property of physical changes explains why matter is conserved in a physical change? The arrangements of particles do not c

hange, and the bonds between atoms do not break. Mass, not energy, changes in a physical change. Energy and mass change in a physical change. The bonds between atoms do not break; it is only the arrangement that changes.
Physics
2 answers:
Irina-Kira [14]3 years ago
6 0

Answer:

d

Explanation:

Gala2k [10]3 years ago
4 0

Answer: Option (d) is the correct answer.

Explanation:

Physical change is defined as a change in which no new product is formed, that is, there is no change in chemical composition of a substance.

For example, when ice melt then solid state of water changes into liquid state.

That is, arrangement of atoms changes and bond between the atoms do no break during a physical change.

Thus, we can conclude that the property bonds between atoms do not break; it is only the arrangement that changes explains why matter is conserved in a physical change.

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If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many imageswou
klasskru [66]

Answer:

If you punched thousands of holes in the aluminum foil, you would see thousands of small images in the viewer.

Explanation:

If thousands of holes is punched/perforated in an aluminum foil, such that there were more holes than foil.

Thousands of small images would be formed, which can be seen in the viewer. Aluminum foil has a good reflective property which allows the thousands of holes punched to reflect equal number of images, although the images are smaller in size than the original image size.

3 0
3 years ago
Vectors A and B lie in the xy ‑plane. Vector A has a magnitude of 17.1 and is at an angle of 150.5∘ counterclockwise from the +x
IRINA_888 [86]

Answer:

A = -14.87 i ^ + 8.42 j ^ + 0 k ^

B = -25.41 i ^ -12.0 j ^ + 0 k ^

Explanation:

For this exercise let's use trigonometry by decomposing to vectors

vector A

module 17.1 with an angle of 150.5 counterclockwise.

         Sin 150.5 = A_{y} / A

         cos 150.5 = Ax / A

         A_{y} = A sin 150.5 = 17.1 sin 150.5

         Aₓ = A cos 1505 = 172 cos 150.5

         A_{y} = 8,420

         Aₓ = -14.870

the vector is

          A = -14.87 i ^ + 8.42 j ^ + 0 k ^

Vector B

         B_{y} = 28.1 sin 205.3

         Bₓ = 28.1 cos 205.3

         B_{y} = -12.009

          Bₓ = -25.405

the vector is

          B = -25.41 i ^ -12.0 j ^ + 0 k ^

5 0
3 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
3 years ago
Near the equator, rising air is associated with a pressure zone known as the _____.
umka2103 [35]
Equatorial low , i think,.. 
6 0
4 years ago
Read 2 more answers
A solid cone is 10 cm high. where is its center of mass?
Ivanshal [37]

Answer:

The center of mass of a cone is located along a line. This line is perpendicular to the base and reaches the apex. The center of mass is a distance 3/4 of the height of the cone with respect to the apex.

Explanation:

3 0
3 years ago
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