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2 years ago

The equations of kinematics describe the motion of an

Physics

1 answer:

Svetradugi [14.3K]2 years ago

3 0

Answer:

The equations of kinematics is applied for the motion with constant acceleration (including zero), but the condition is that the acceleration should be in the direction of the motion (positive or negative).

In circular motion, the acceleration is radial (centripetal), which means that the acceleration is always perpendicular to the motion of the object, therefore the equations of kinematics cannot be applied.

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A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a

Stella [2.4K]

Answer:

y = 67.6 feet, y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

tan θ = y / L

For the small triangle. Projector up to the person

tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

y₀ / (L -d) = y / L

y = y₀ L / (L-d)

The distance from the screen (d), we look for it with kinematics

v = d / t

d = v t

we replace

y = y₀ L / (L - v t)

y = 5.2 22 / (22 - 3 t)

y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

y = 114.4 / (22-13)

y = 67.6 feet

7 0

3 years ago

What happens to the coefficient of friction when the weight is increased? Why is this?

Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0

3 years ago

Read 2 more answers

If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the tot

zheka24 [161]

The portion of the flux leaves the curved surface of the cylinder is 60%.

<h3 /><h3>What are electrons?</h3>

The electrons are the spinning objects around the nucleus of the atom of the element in an orbit.

If a point charge is located at the center of a cylinder and the electric flux leaving one end of the cylinder is 20% of the total flux leaving the cylinder.

If 20% of the flux leave from one end, then another 20% will leave from another end.

So, the net flux through curved surface is

100 -20 -20 = 60%

Thus, the total flux leaves the curved surface of the cylinder is 60%

Learn more about electrons.

brainly.com/question/1255220

#SPJ1

5 0

1 year ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago

An electric wheel chair is designed to run on a single 12-V battery rated to provide 100 ampere-hours. How much energy is stored

Paha777 [63]

Explanation:

It is given that,

Voltage of the battery, V = 12 V

Current, I = 100 ampere-hours

Energy stored is given by the product of power and time taken. So,

$E=P\times t$

P is the power, $P=V\times I$

$P=12\times 100$

P = 1200 watts

This power can be used for 1 hour or 3600 seconds

Energy, $E=1200\times 3600$

E = 4320000 J

So, the energy stored in this battery is 4320000 J. Hence, this is the required solution.

5 0

3 years ago