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3 years ago

A certain wave has a compressions and rarefactions.How should this wave be classified?

Physics

1 answer:

AlexFokin [52]3 years ago

7 0

A) as a longitudinal wave

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a 5.00 kg plane is speeding up on the ground with an applied force of 706 N. If the net force is 450 N, what would be the force

loris [4]

Answer:

The force due to air resistance is 256 N.

Explanation:

Given;

mass of the plane, m = 5 kg

applied force on the plane, Fa = 706 N

the net force on the plane, ∑F= 450 N

Let the force due to air resistance = Fr

The net force on the plane is given as;

Net force = applied force - force due to air resistance

∑F = Fa - Fr

Fr = Fa - ∑F

Fr = 706 - 450

Fr = 256 N.

Therefore, the force due to air resistance is 256 N.

5 0

3 years ago

A hurricane becomes more powerful by evaporating water from A) lakes. B) mountains. C) oceans. D) rivers.

Luda [366]

A hurricane becomes more powerful by evaporating water from oceans

3 0

3 years ago

Read 2 more answers

Which statement is true if the refractive index of medium A is greater than that of medium B?

allsm [11]

Answer:

I think it is C

Explanation:

7 0

3 years ago

The pressure of a gas is 15 atm in a 5L cylinder. If the volume of the cylinder is depressed to 3L, What is the new pressure exe

disa [49]

P1v1=p2v2. 15x5=p2x3

5 0

3 years ago

A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, $\lambda = 1\ nC = 1\times 10^{- 9}\ C$

Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

$\lambda$ = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

7 0

3 years ago