All categories

4 years ago

Identify a wave that speeds up when it passes from air to water as well as one that slows down

Physics

1 answer:

diamong [38]4 years ago

3 0

Answer:

A sound wave speeds up as it passes from air to water. A light wave slows down as it passes from air to water.

Explanation:

You might be interested in

When matter transitions between solid liquid and gas

iVinArrow [24]

Its called the Phase Transition. Im assuming your asking what is it called when matter transitions between solid liquid and gas.

5 0

3 years ago

Label each formula in the chemical quation below as either a reactant of a product.

Karolina [17]

Fe and S are both reactants: they react with each other to give a different compound.

FeS is the product of the reaction: it was formed or produced as result of the reaction of Fe and S.

Answer:

Fe: reactant
S: reactant
FeS: product

3 0

4 years ago

A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0

IgorLugansk [536]

Answer:

Magnitude of the force is

$F_3 = 2.83$

direction of the force is given as

$\theta = 45 degree$ West of South

Explanation:

As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy

So here we have

$\vec F_1 + \vec F_2 + \vec F_3 = 0$

here we know that first force is of magnitude 2 N towards east

$\vec F_1 = 2 \hat i N$

second force is also of 2.0 N due North

$\vec F_2 = 2 \hat j$

now from above equation

$2\hat i + 2\hat j + \vec F_3 = 0$

$\vec F_3 = -2\hat i - 2\hat j$

so magnitude of the force is given as

$F_3 = \sqrt{2^2 + 2^2}$

$F_3 = 2.83$

direction of the force is given as

$\theta = tan^{-1}\frac{F_y}{F_x}$

$\theta = tan^{-1}\frac{-2}{-2}$

$\theta = 45 degree$ West of South

3 0

3 years ago

Read 2 more answers

A 108 kg clock initially at rest on a horizontal floor requires a 653 N horizontal force to set it in motion. After the clock is

kifflom [539]

Answer:

$\mu_s=0.61$

$\mu_k=0.49$

Explanation:

Given that,

Mass of the clock, m = 108 kg

When the clock is not moving, force acting on it, $F_1=653\ N$

For the clock in motion, force acting on it, $F_2=527\ N$

To find,

$\mu_s\ and\ \mu_k$

Solution,

When an object is at rest, the force acting on it is called force due to static friction and if the object is in motion, the force acting on its called force due to kinetic friction.

Let $\mu_s$ is the coefficient of static friction. Force is given by :