A satellite camera in space took this picture of northwestern Algeria, showing an impact crater, sedimentary

Sound travels through a

attashe74 [19]

Answer:

A.3.64 m

Explanation:

Because

v=(fλ)
(1382)=(380)λ
λ=3.637m~3.64m

where ,v=velocity

f=frequency 

λ=wave length 

8 0

1 year ago

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A device for training astronauts and jet fighter pilots is designed to rotate the trainee in a horizontal circle of radius 11.0

kvv77 [185]

The velocity of the trainee is 29 m/s or 0.42 rev/s

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Centripetal Acceleration of circular motion could be calculated using following formula:

$\large {\boxed {a_s = v^2 / R} }$

a = centripetal acceleration ( m/s² )

v = velocity ( m/s )

R = radius of circle ( m )

Let us now tackle the problem!

Given:

Radius of horizontal circle = R = 11.0 m

Force Felt by the Trainee = F = 7.80w

Unknown:

Velocity of Rotation = v = ?

Solution:

$F = ma$

$F = m\frac{v^2}{R}$

$7.80w = m\frac{v^2}{R}$

$7.80mg = m\frac{v^2}{R}$

$7.80g = \frac{v^2}{R}$

$7.80 \times 9.8 = \frac{v^2}{11.0}$

$v^2 = 840.84$

$v \approx 29 ~m/s$

$\omega = \frac{v}{R}$ → in rad/s

$\omega = \frac{v}{2 \pi R}$ → in rev/s

$\omega = \frac{29}{2 \pi \times 11.0}$

$\omega \approx 0.42 ~ rev/s$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Uniform Circular Motion : brainly.com/question/2562955
Trajectory Motion : brainly.com/question/8656387

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate , Circular , Ball , Centripetal

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3 years ago

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a rifle is fired and recoils when the bullet leaves the gun. this is an example of newton's 3rd law. the force on the bullet is.

Alex_Xolod [135]

The force of the bullet is the same.

3 0

2 years ago

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CD bisects AB at point G. If AE = BE, which equation must be true?

ratelena [41]

AG = BG equation must be true

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3 years ago

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A particle moving at speed 0. 32 c has momentum p0. the speed of the particle is increased to 0. 72 c. what is its momentum now?

Semenov [28]

A particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c then its momentum would be 2.25P₀.

<h3>What is momentum?</h3>

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

Mathematically the formula of the momentum is

P = mv

where P is the momentum of the particle

m is the mass of the particle

v is the velocity by which the particle is moving

As given in the question a particle moving at speed 0.32 c has momentum P₀. the speed of the particle is increased to 0.72 c

by using the formula of the momentum and substituting the values of the velocity

P =mv

As mentioned in the question when the particle is moving with 0.32c velocity it has a momentum of P₀

P₀ = m*(0.32c)

If the speed of the particle is increased to 0.72 c the momentum would be

P=m*(0.72c)

by dividing the second equation from the first one

P₀/P = 0.32c/0.72c

P₀/P = 0.44

P =2.25P₀

Thus, the increased momentum of the particle would be 2.25P₀

Learn more about momentum from here

brainly.com/question/17662202

#SPJ4

6 0

1 year ago