When a wave is bent by traveling from one medium to another

Explain the difference between science usage and common usage for the term resistance

san4es73 [151]

Keremiad<span> is a long literary work, usually in prose, but sometimes in verse, in which the author bitterly laments the state of society and its morals in a serious tone of sustained invective, and always contains a prophecy of society's imminent downfall. </span>

3 0

2 years ago

Pls help it’s easy ik but I want to make sure of them both I don’t have much time

LuckyWell [14K]

Answer:

To the right

Weight

Explanation:

8 0

1 year ago

On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around

Mekhanik [1.2K]

Answer:

$v\approx 16.956\,\frac{m}{s}$

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

$\Sigma F = f = m\cdot \frac{v^{2}}{R}$

The maximum speed is:

$v = \sqrt{\frac{f\cdot R}{m} }$

$v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }$

$v\approx 16.956\,\frac{m}{s}$

7 0

3 years ago

A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.

SVEN [57.7K]

Each side has to have at least 44 horses

F61160 N. This is further explained below.

<h3>What is the force?</h3>

Generally, We are only interested in the component that operates horizontally since the vertical components all cancel each other out. The pressure difference works on the hemisphere to generate a normal force all over the surface, but we are only concerned with that force's horizontal component. This may be determined by supposing the hemispheres to be two flat circular plates of the same radius as the hemispheres that have been forced together.

Therefore, force is equal to pressure multiplied by area, which is

F= (970 -15 )( * (0.45 m)2)

F=60754 N for each side.

Therefore, each side has to have at least 44 horses

44* 1390 = 61160 N

Read more about force

brainly.com/question/13191643

#SPJ1

3 0

1 year ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

2 years ago