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4 years ago

There is a reflection at the boundary between two materials if there is a change of __________ at the boundary.

Physics

1 answer:

melisa1 [442]4 years ago

8 0

<span>The speed of the light in the materials. Hope this helps!!</span>

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The pulley is a uniform disk with a mass of 0.75 kg and a radius of 6.5

larisa [96]

Answer:

please find the answer in the attached images , in 2 parts, btw thanks for so many points. i hope my answers are correct.

5 0

3 years ago

Read 2 more answers

An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its

FromTheMoon [43]

Answer:

The terminal speed of this object is 12.6 m/s

Explanation:

It is given that,

Mass of the object, m = 80 kg

The magnitude of drag force is,

$F_{drag}=12v+4v^2$

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.

$F_{drag}=mg$

$12v+4v^2=80\times 9.8$

$4v^2+12v=784$

On solving the above quadratic equation, we get two values of v as :

v = 12.58 m/s

v = -15.58 m/s (not possible)

So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.

6 0

3 years ago

3. A 40-gram ball of clay is dropped from a height, h, above a cup which is attached to a spring of spring force constant, k, of

zheka24 [161]

Answer:

the maximum speed of the ball is 12.65 m/s

Explanation:

Given;

mass of the ball, m = 40 g = 0.04 kg

spring constant, k = 25 N/m

Apply the principle of conservation of energy;

The Elastic potential energy of the spring will be converted into Kinetic of the ball;

$\frac{1}{2} kx^2 = \frac{1}{2} mv^2\\\\ kx^2 = mv^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m}} \\\\v = \sqrt{\frac{(25)(0.506)^2}{0.04}} \\\\v = \sqrt{160.0225} \\\\v = 12.65 \ m/s$

Therefore, the maximum speed of the ball is 12.65 m/s

4 0

3 years ago

The motion of an electron is given by x(t)=pt3+qt2+r, with p = -2.3 m/s^3 ,q = +1.5 m/s^2 , and r = +9.0 m.A) Determine its velo

alexandr402 [8]

Answer:

$v(0)=0\\v(1)=-3.9\ m/s\\v(2)=-21.6\ m/s\\v(3)=-53.1\ m/s$

Explanation:

<u>Instant Velocity </u>

Given the position as a function of time x(t), the instant velocity is the derivative of the function:

$v(t)=x'(t)$

We are given the position as

$x(t)=-2.3t^3+1.5t^2+9$

The derivative of x is

$v(t)=x'(t)=-6.9t^2+3.0t$

A) Let's compute v(0)

$v(0)=-6.9(0)^2+3.0(0)=0$

$v(1)=-6.9(1)^2+3.0(1)$

$v(1)=-3.9\ m/s$

$v(2)=-6.9(2)^2+3.0(2)$

$v(2)=-21.6\ m/s$

$v(3)=-6.9(3)^2+3.0(3)$

$v(3)=-53.1\ m/s$

4 0

3 years ago

BRAINLIEST!!! how does the electric force between two charged particles change if the distance between them is reduced by a fact

Bogdan [553]

Answer:

F = K Q1 * Q2 / R^2 force between 2 charged particles

F2 / F1 = R1^2 / R2^2 = (R1 / R2)^2 = (R1 / R1 /2)^2 = 2^2 = 4

d) the force would be increased by a factor of 4

3 0

2 years ago