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4 years ago

Which of the following observations represent conclusive evidence of an interaction? (Select all that apply.)

Physics

1 answer:

9966 [12]4 years ago

6 0

Answer:

a,b,d and e are correct.

Explanation:

a) Change of temperature without change of velocity is a conclusive evidence of an interaction. As the temperature of a body will only change if heat energy is flows in or out of the body with surrounding. So, option a) is correct.

b) Change of the direction without change of speed is an evidence of an interaction. The direction changes when the object is under a perpendicular acceleration acceleration which clearly means the particle is interacting or external force applied. So, option b is correct.

d) Change of shape or configuration without change of velocity is conclusive evidence of an interaction. The shape can change only if external stress is applied on the particle. So, Option d is correct.

e) Change of identity without change of velocity is a conclusive evidence of an interaction. The identity of an object can only change if it interacts with surroundings. So, option e is correct

Therefore the only incorrect option is c .

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3 years ago

Clara rushes 30m to a truck then turns and walks back. Total travel is 120s what is her average velocity?

navik [9.2K]

Taking into account the definition of velocity, Clara's average velocity is 0.5 m/s.

<h3>Definition of velocity</h3>

Velocity is a physical magnitude that relates the displacement of an object, the time it takes to make this change in position and direction. So it is considered a vector magnitude.

In other words, the velocity can be defined as the amount of space traveled per unit of time with which a body moves, considering the direction, and can be calculated using the expression:

velocity= distance traveled÷ time

<h3>Average velocity of Clara</h3>

Clara rushes 30 m to a truck then turns and walks back. Total travel is 120s. Then, you know:

distance traveled= 30m rushing + 30m walking back= 60 m
time= 120 s

Replacing in the definition of velocity:

velocity= 60 m÷ 120 s

Solving:

<u><em>velocity= 0.5 m/s</em></u>

Finally, Clara's average velocity is 0.5 m/s.

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2 years ago

An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance

Lorico [155]

Answer:

The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.

Explanation:

Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

$F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}$

and since the electric field E in between parallel plates separated a distance d and under a potential difference $\Delta V$ , is given by:

$E=\frac{\Delta\,V}{d}$

then :

$a=\frac{q\,\Delta V}{m\,d}$

We want to find when the particle reaches velocity zero via kinematics:

$v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a$

We replace this time (t) in the kinematic equation for the particle displacement:

$\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}$

Replacing the values with the information given, converting the distance d into meters (0.01 m), using $\Delta V=100\,V$ , and the electron's kinetic energy:

$\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J$

we get:

$\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters$ Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:

0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]

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4 years ago

Find the area under the standard normal curve between z=0.19 and z=2.18. round your answer to four decimal places, if necessary.

LiRa [457]

The area under the standard normal curve between z=0.19 and z=2.18 is,

P(0.19<z<2.18)= 0.4101

<h3>How can we calculate the area under the curve?</h3>

To calculate the The area under the standard normal curve between z=0.19 and z=2.18, we are using two things,

<u>Step 1</u>: The formula,

P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)

<u>Step 2</u>: The statistical values of the area under the curve we get from the picture.

Now we put the known values from the picture in the above formula, we get,

P(0.19<z<2.18)= P(Z<2.18)- P(Z<0.19)

Or, P(0.19<z<2.18)= 0.9854-0.5753

Or, P(0.19<z<2.18)= 0.4101

From the above calculation we can easily conclude that,

The area under the standard normal curve between z=0.19 and z=2.18 is,

P(0.19<z<2.18)= 0.4101

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2 years ago