Ask question

Ask question

All categories

4 years ago

11

What type of waves move energy forward, but the source moves up and down?

1 answer:

Tresset [83]4 years ago

3 0

Transverse waves is the answer.
Hope I helped if you need anything else ask me

You might be interested in

During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa

maw [93]

Answer:

Explanation:

Given

Initial speed $u=60\ mi/hr\approx 88\ ft/s$

distance traveled before coming to rest $d_1=120\ ft$

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$0-(88)^2=2\times a\times 120---1$

for $u_2=80\ mi/hr\approx 117.33\ ft/s$

using same relation we get

$0-(117.33)^2=2\times a\times (d_2)----2$

divide 1 and 2 we get

$(\frac{88}{117.33})^2=\frac{120}{d_2}$

$d_2=213.32\ ft$

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0

3 years ago

Jose is creating a game for class. In his game, fellow students will have to identify different types of communities. Which of t

stich3 [128]

Answer:

Citizens and neighborhood.

3 0

3 years ago

Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th

Evgen [1.6K]

Answer:

$\omega=0.24\ rad.s^{-1}$

Explanation:

Given:

mass of person, $m=36\ kg$

mass of merry go-round, $M=300\ kg$

radius of merry go-round, $R=2\ m$

velocity of the person running, $v=4\ m.s^{-1}$

<u>We consider merry go-round as a ring:</u>

Now the moment of inertial of the ring is given as,

$I=M.R^2$

$I=300\times 2^2$

$I=1200\ kg.m^{-2}$

<u>Moment of inertia of the person considering as a point mass:</u>

$I_p=m.R^2$

$I_p=36\times 2^2$

$I_p=144\ kg.m^2$

<u>Now according to the conservation of angular momentum:</u>

$I.\omega=I_p.\omega_p$

where:

$\omega =$ angular velocity of the merry-go-round

$\omega_p=$ angular velocity of the person running

$1200\times \omega=144\times \frac{v}{R}$

$\omega=\frac{144}{1200} \times \frac{4}{2}$

$\omega=0.24\ rad.s^{-1}$

4 0

4 years ago

Read 2 more answers

The filament of an electric bulb draws a current of 0.4 ampere. Calculate the amount of charge flowing through the filament if t

Marina86 [1]

Here I =0.4 amp and t=2hr or 2×60×60 sec
We known I=nQ/t where Q is charge and n is no of charge.
n=It/Q
0.4x(2x60x60)/1.6x10^-19

3 0

3 years ago

How much time should be allowed for a 605-mile car trip if the car will the traveling at an average speed of 55 Miles per hour?

Luda [366]

T = d/s

T = 605/55

T = 11

Hope that helps!!

3 0

3 years ago

Read 2 more answers