All categories

3 years ago

In 3.0 seconds, a wave generator produces 15 pulses that spread over a distance of 45 centimeters. Calculate its period.

Physics

1 answer:

melisa1 [442]3 years ago

4 0

Period = (time) / (number of pulses)

= (3 sec) / (15 pulses)

= 0.2 sec/pulse

You might be interested in

A car with a mass of 850kg is moving at a speed of 72km/h when colliding with a concrete wall until it stops. After the collisio

Sergeeva-Olga [200]

Answer:

Explanation:

The vehicle is experiencing a large force created by the concrete wall.

Equation

vf^2 = vi^2 + 2*a * d

Givens

vf = 0 The car eventually does stop.

vi = 72 km/hr * [ 1000 m/ km] * [1 hour / 3600 seconds]

vi = 20 meters / second

a = ?

m = 850 kg

Solution

vf^2 = vi^2 + 2a*d

0 = 20 m/s + 2* 2 *a

-20 m/s = 4a

-20/4 = a

a = - 5 m/s^2 The minus sign tells you the vehicle is slowing down. It sure should be.

Force = m * a

F = - 850 * (-5)

F = - 4250 N

The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.

8 0

3 years ago

PLEASE HELP ME WITH THIS!!!!<br><br><br>

vodka [1.7K]

Answer:

wasts materials such as cans. it can hit space crafts and satellites. no one has ever been killed by space debris, space vessels very rarely are seriously damaged.

7 0

3 years ago

The maximum stress in a section of a circular tube subject to a torque is τmax = 27 MPa . If the inner diameter is Di = 3.75 cm

DIA [1.3K]

Answer:

$T_{max} = 4.735\,kN\cdot m$

Explanation:

The shear stress due to torque can be calculed by using the following model:

$\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}$

The maximum torque on the section is:

$T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}$

The Torsion Constant for the circular tube is:

$J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})$

$J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]$

$J_{tube} = 4.560\times 10^{-6}\,m^{4}$

Now, the require output is computed:

$T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}$

$T_{max} = 4.735\,kN\cdot m$

7 0

3 years ago

Read 2 more answers

What is the weight on mass

Gre4nikov [31]

Answer:

6.39×10^23 kg is the weight on mass

3 0

2 years ago

x rays with wavelength of 2.0nm scatter from a nacl crystal with plane spacing 0.281 nm find the scattering

ValentinkaMS [17]

Explanation:

It is given that,

Wavelength of x-rays = 2 nm

Plane spacing, d = 0.281 nm

It is assumed to find the scattering angle for second order maxima.

For 2nd order, Bragg's law is given by :

$2d\sin\theta=n\lambda$

For second order, n = 2

$\sin\theta=\dfrac{n\lambda}{2d}\\\\\sin\theta=\dfrac{2\times 2\ nm}{2\times 0.28\ nm}\\\\\theta=\sin^{-1}(7.14)$

Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.

4 0

3 years ago