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1 year ago

The speed of sound in air is 340 m/s. What is the wavelength of a soundwave that has a frequency of 502 Hz?

Physics

1 answer:

Mama L [17]1 year ago

5 0

The formula for calculating velocity of wave is expressed as

v = fλ

where

v = velocity of the wave

λ = wavelength

f = frequency of the wave

From the information given,

v = 340

f = 502

Substituting these values into the formula, we have

340 = 502λ

Dividing both sides of the equation by 502,

340/502 = 502λ/502

λ = 0.68 m

The wavelength is 0.68 m

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A 25N force is acting on a body moving on a straight line with Initial momentum 20 kam's. Find the final momentum after 4 second

Nezavi [6.7K]

The final momentum of the body is equal to 120 Kg.m/s.

<h3>What is momentum?</h3>

Momentum can be described as the multiplication of the mass and velocity of an object. Momentum is a vector quantity as it carries magnitude and direction.

If m is an object's mass and v is its velocity then the object's momentum p is: $\mathbf {p} =m\mathbf {v$ . The S.I. unit of measurement of momentum is kg⋅m/s, which is equivalent to the N.s.

Given the initial momentum of the body = Pi = 20 Kg.m/s

The force acting on the body, Pf = 25 N

The time, Δt = 4-0 = 4s

The Force is equal to the change in momentum: F ×Δt = ΔP

25 × 4 = P - 20

100 = P - 20

P = 100 + 20 = 120 Kg.m/s

Therefore, the final momentum of a body is 120 Kg.m/s.

Learn more about momentum, here:

brainly.com/question/4956182

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7 months ago

A rightward force of 12.0 N is applied to a 2.0-kg object to accelerate it across a horizontal

ad-work [718]

Answer:

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Explanation:

Net accelerating force becomes 12-8 = 4 N

F = ma

4 = 2 * a

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1 year ago

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Department of Highway Safety and Motor Vehicles OR

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3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

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2 years ago

A 0.68 kg squirrel is resting on a branch 8 meters above the ground. What is the gravitational potential energy of a squirrel? A

ANEK [815]

Answer:

The gravitational potential energy of a squirrel is 53.312 J.

Explanation:

We have,

Mass of a squirrel is 0.68 kg

It is placed at a height of 8 m above the ground.

It is required to find the gravitational potential energy of a squirrel. It is possessed by an object due to its position. Its formula is given by :

$E=mgh\\\\E=0.68\times 9.8\times 8\\\\E=53.312\ J$

So, the gravitational potential energy of a squirrel is 53.312 J.

7 0

2 years ago