If you apply a force of 100 N to the level, how much force is applied to lift the crate?

A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values

Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or $\lambda=20.0cm$

c. The period, or T, of a wave is found in the equation

$f=\frac{1}{T}$ were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

$50.0=\frac{1}{T}$ so

$T=\frac{1}{50.0}$ and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

$f=\frac{v}{\lambda}$ and since we already have the frequency and we solved for the wavelength already, filling in:

$50.0=\frac{v}{20.0}$ and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

5 0

3 years ago

The area of the pond is approximately equal to the area of a circle with radius 297m. Find the mass of the ice. Answer in kilogr

True [87]

Answer:

mass of the ice is 254980463.8T kg

where T is the value of the thickness omitted in the question.

Explanation:

The ice on Walden Pond is .......... thick. The area of the pond is approximately equal to the area of a circle with radius 297 m. Find the mass of the ice. Answer in kg.

The value of the thickness of the ice T is omitted, but I will show the solution, and the real answer can be gotten by multiplying the final calculated answer here by the thickness of the ice omitted.

Given the radius of the equivalent circle of the ice = 297 m'

the area of the ice can be gotten from area A = $\pi r^{2}$ = $3.142*297^{2}$ = 277152.678 m^2

recall that the density of ice p ≅ 920 kg/m^3

also,

density of ice p = (mass of ice, m) ÷ (volume of ice, v)

i.e p = m/v

and,

m = pv

substituting the value of the density of water p into the equation, we have,

mass of the ice, m = 920v ....... equ 1

The volume of the ice above will be = (area of the ice, A) x (thickness of the ice, T)

i.e v = AT

substituting the value of area A into the equation, we have

v = 277152.678T ......equ 2

substitute value of v into equ 1

mass of the ice, m = 920 x (277152.678T)

mass of the ice, m = 254980463.8T kg

where T is the thickness of the ice

NB: To get the mass, multiply this answer with the thickness T given in the question.

7 0

3 years ago

A bullet of mass 0.0021 kg initially moving at 497 m/s embeds itself in a large fixed piece of wood and travels 0.65 m before co

alina1380 [7]

Answer:

The force exerted by the wood on the bullet is 399.01 N

Explanation:

Given;

mass of bullet, m = 0.0021 kg

initial velocity of the bullet, u = 497 m/s

final velocity of the bullet, v = 0

distance traveled by the bullet, S = 0.65 m

Determine the acceleration of the bullet which is the deceleration.

Apply kinematic equation;

V² = U² + 2aS

0 = 497² - (2 x 0.65)a

0 = 247009 - 1.3a

1.3a = 247009

a = 247009 / 1.3

a = 190006.92 m/s²

Finally, apply Newton's second law of motion to determine the force exerted by the wood on the bullet;

F = ma

F = 0.0021 x 190006.92

F = 399.01 N

Therefore, the force exerted by the wood on the bullet is 399.01 N

6 0

3 years ago

Drag and drop each description into the appropriate category:

Anestetic [448]

Answer:

All of the following are already correctly categorized. This facts about these radioactive decays are true. Further, I have explained below in explanation portion.

Explanation:

Alpha Decay: Yes it cannot penetrate through paper & Yes it emits Alpha particle when an element undergoes the process of Alpha Decay. Here's how:

First of all, we need to understand why a particular element decays? why radioactive decay occurs?

It is because, everything in this universe tends to achieve its stable state. Everything put efforts to be in their stable state. Likewise, when radioactive elements like Uranium or Thorium become unstable they undergo the process of Alpha decay. As a result, they emit energy and alpha particle.

Furthermore, Alpha decay has low penetration power and it has higher mass than Beta Decay and Gamma Decay that's why it cannot travel in long distances. Hence, it cannot pass through a sheet of paper.

Beta Decay: Yes it emits electrons or positrons and Yes it can penetrate through paper but cannot penetrate through a foil. Here's how:

Till now, we are familiar that why the process of decay occurs right? in order to be stable. Like wise Beta Decay occurs and it emits beta particle and energy. The main difference between alpha decay and beta decay is that alpha particle's mass is much greater than beta particle's mass because beta particle is either a negatively charged electron or positron (a positively charged electron or anti-electron) and we know that mass electron is much smaller.

So, Beta Decay has moderate penetrate power and it's mass is much smaller that's why it can travel in longer distances and it can penetrate paper but cannot penetrate into foil.

Gamma Decay: Yes it requires thick lead shield and yes it emits photon. Here's how:

Gamma Decay is a radioactive wave, it only emits energy in form of photons and it has no mass and no charge and but it has very low wavelength but highest energy with great penetration power and that is the reason that it can penetrate paper, it can penetrate foil but it cannot penetrate thick lead shield.

Hope, this will help :)

6 0

3 years ago

Read 2 more answers

In the chemical formula 3NH4, how many total atoms of nitrogen (N

Rus_ich [418]

One atom on nitrogen because there isn’t a number directly after it’s the 3 signifies that there is three groups of those atoms as a total

8 0

3 years ago

Read 2 more answers