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Ksivusya [100]
3 years ago
7

What happens when a substance undergoes a physical change?

Physics
2 answers:
Andreyy893 years ago
4 0
It visually changes but keeps it's properties like ice to water to steam.
zaharov [31]3 years ago
4 0

Answer:

Its form is affected

Explanation:

A physical change implies only a modification in the physical properties of a substance. Contrary to this in a chemical change not only its form but its composition changes.

Processes like melting, transition to a gas, change of strength, textural change, shape, size, color, volume, and density, are examples of physical changes.

You might be interested in
An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea
german

Answer:

Q1)  Time taking by particle to travel the 40 km wrt. earth = 1.34\times10^{-6} sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = 1.285\times10^{-5} sec.

Explanation:

Given :

Speed of particle wrt. earth v=0.99537c

Distance between where particle is created and earth surface = 40 km

we know that,

⇒       v = \frac{x}{t}

Where x = 40\times10^{3} m, v = 0.99537c, we know speed of light c = 3 \times10^{8}

∴      t = \frac{x}{v}

         = \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }

      t = 1.34\times10^{-6} sec

∴ Thus, time taking by particle to travel the 40 km wrt. earth t = 1.34\times10^{-6} sec

According to the lorentz transformation,

⇒    l = l_{o} \sqrt{1-\frac{v^2}{c^2} }

Where l = improper length, l_{o} =proper length (distance measured wrt. rest frame) = 40 km

     l = 40 \sqrt{1-\frac{v^2}{c^2 }

     l = 40 \times 0.096

     l = 3.84 km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

   \Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }

Where \Delta t = improper time (wrt. earth frame time) =1.34\times10^{-6} sec ,  \Delta t _{o} = proper time (wrt. particle frame).

 1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}

 \Delta t_{o} = 1.285 \times10^{-5} sec

Thus, the time taking by particle to travel from where it is created to the surface of the earth = 1.285 \times10^{-5} sec.

6 0
3 years ago
Which of the following statements is/are true?Check all that apply.a. A conservative force permits a two-way conversion between
Eddi Din [679]

Answer:

a). A conservative force permits a two-way conversion between kinetic and potential energies.

TRUE

Because there is no energy loss in presence of conservative forces so energy conversion in two ways are possible.

b). A potential energy function can be specified for a conservative force.

TRUE

negative gradient of potential energy is equal to conservative force

F = -\frac{dU}{dr}

c). A non-conservative force permits a two-way conversion between kinetic and potential energies.

FALSE

here energy is lost against non-conservative forces

d). The work done by a conservative force depends on the path taken.

FALSE

work done by conservative force is independent of path

e). The work done by a non-conservative force depends on the path taken.

TRUE

work done by non conservative forces depends on path.

f). A potential energy function can be specified for a non-conservative force.

FALSE

It is not defined for non conservative forces

3 0
3 years ago
Two obiect accumulated a charge of
tamaranim1 [39]

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

q_1=4.5\mu C=4.5\cdot 10^{-6}C is the magnitude of the 1st charge

q_2=2.8\mu C=2.8\cdot 10^{-6}C is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N

So, the closest answer is

A) 181.24 N

3 0
3 years ago
A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force
salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J

Energy lost due to friction

W=Fd=0.031\times 0.158=0.0048J

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

\frac{1}{2}mv^2=0.0052

\frac{1}{2}\times 0.00524\times v^2=0.0052

v = 1.40 m/sec

7 0
3 years ago
Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo
777dan777 [17]

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert  42 m/s ⇒   km/h

1 km =1000 m

1 h = 36000 sec

42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\

0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}

Step 2

find kilometers traveled after 4  hours

V=\frac{s}{t}\\ \\

V,velocity

s, distance traveled

t. time

now, isolating s

V=\frac{s}{t} \\s=V * t\\

and replacing

s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\

S=604.8 Km

Have a great day

4 0
3 years ago
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