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3 years ago

Which of the following is true regarding transverse waves?

Physics

1 answer:

DIA [1.3K]3 years ago

4 0

Answer:

in a transverse wave, The directions that energy and matter travel in are perpendicular to one another.

these waves are types of mechanical waves that doesn't require any material medium to transfer.

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A compound consisting of only phosphorous and oxygen atoms is 43.64 % phosphorus by mass.The molecular mass/formula weight of th

DanielleElmas [232]

First, we calculate the mass of Phosphorous present:
283.88 x 0.4364
= 123.88 amu
Atomic mass of P is 31 amu

moles of P = mass / Ar
= 123.88 / 31
= 4.0 moles

We know that one mole of substance has 6.02 x 10²³ particles
Atoms of P = 4 x 6.02 x 10²³
= 2.41 x 10²⁴ atoms

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3 years ago

Como foi a despedida do bilhete?

Dmitrij [34]

Can you translate to english ?

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3 years ago

He power output of a cyclist moving at a constant speed of 5.3 m/s on a level road is 120 w. (a) what is the force exerted on th

EleoNora [17]

P=F*V
F=P/V
F=120/5.3
F=22.6N

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3 years ago

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid

nataly862011 [7]

Answer:

the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

data that were not necessary to the solution are;

a) mass of truck and b) mass of load

Explanation:

Given that;

mass of load $m_{LS}$ = 10000 kg

mass of flat bed $m_{FB}$ = 20000 kg

initial speed of truck $v_{0}$ = 12 m/s

coefficient of friction between the load sits and flat bed μs = 0.5

A) the minimum stopping distance for which the load will not slide forward relative to the truck.

Now, using the expression

Fs,max = μs $F_{N}$ -------------let this be equation 1

where $F_{N}$ = normal force = mg

Fs,max = μs mg

ma $_{max}$ = μs mg

divide through by mass

a $_{max}$ = μs g ---------- let this be equation 2

in equation 2, we substitute in our values

a $_{max}$ = 0.5 × 9.8 m/s²

a $_{max}$ = 4.9 m/s²

now, from the third equation of motion

v² = u² + 2as

$v_{f}$ ² = $v_{0}$ ² + 2aΔx

where $v_{f}$ is final velocity ( 0 m/s )

a is acceleration( - 4.9 m/s² )

so we substitute

(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx

0 = 144 m²/s² - 9.8 m/s²Δx

9.8 m/s²Δx = 144 m²/s²

Δx = 144 m²/s² / 9.8 m/s²

Δx = 14 m

Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m

B) data that were not necessary to the solution are;

a) mass of truck and b) mass of load

3 0

3 years ago

Four 50-g point masses are at the corners of a square with 20-cm sides. What is the moment of inertia of this system about an ax

deff fn [24]

Answer:

moment of inertia I ≈ 4.0 x 10⁻³ kg.m²

Explanation:

given

point masses = 50g = 0.050kg

note: m₁=m₂=m₃=m₄=50g = 0.050kg

distance, r, from masses to eachother = 20cm = 0.20m

the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by

= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m

moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center

mathematically,

I = ∑m×d²

remember, a square will have 4 equal points

I = ∑m×d² = 4(m×d²)

I = 4 × 0.050 × (14.12 x 10⁻² m)²

I = 0.20 × 1.96 × 10⁻²

I = 3.92 x 10⁻³ kg.m²

I ≈ 4.0 x 10⁻³ kg.m²

attached is the diagram of the equation

7 0

3 years ago