The weight of a body in earth is 29N. Find the mass?

What is the difference between transverse and longitudinal waves?

Fed [463]

To explain how transverse and longitudinal waves work, let us give two examples for each particular case.

In the case of transverse waves, the displacement of the medium is PERPENDICULAR to the direction of the wave. One way to visualize this effect is when you have a rope and between two people the rope is shaken horizontally. The shift is done from top to bottom. This phenomenon is common to see it in solids but rarely in liquids and gases. A common application usually occurs in electromagnetic radiation.

On the other hand in the longitudinal waves the displacement of the medium is PARALLEL to the direction of propagation of the wave. A clear example of this phenomenon is when a Slinky is pushed along a table where each of the rings will also move. From practice, sound waves enclose the definition of longitudinal wave displacement.

Therefore the correct answer is:

C. In transverse waves the displacement is perpendicular to the direction of propagation of the wave, while in longitudinal waves the displacement is parallel to the direction of propagation.

8 0

4 years ago

Determine the pressure P of a 450K Oxygen gas in a gas chamber when its initial pressure is 175Pa has a temp 300K

FrozenT [24]

Answer:

262.5Pa

Explanation:

Given parameters:

P1 = 175Pa

T1 = 300K

T2 = 450K

Unknown:

P2 = ?

Solution:

To solve this problem, we are going to use an adaptation of the combined gas law.

$\frac{P1}{T1}$ = $\frac{P2}{T2}$

Insert the parameters and solve;

$\frac{175}{300}$ = $\frac{P2}{450}$

P2 = 262.5Pa

8 0

3 years ago

(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you ca

Colt1911 [192]

a) 52.5 m

b) 16.0 m/s

Explanation:

a)

The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:

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

where

v = 0 is the final velocity (the car comes to a stop)

u = 28.7 m/s is the initial velocity of the car

a is the acceleration

s is the stopping distance

For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:

$a=\frac{-\mu mg}{m}=-\mu g$

where:

$\mu=0.80$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting and solving for s, we find:

$s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m$

b)

In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is

$\mu=0.25$

Here we want the stopping distance of the car to remain the same as part a), so

$s=52.5 m$

We can use again the same suvat equation:

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

And since the final velocity is zero

u = 0

We can find the initial velocity of the car:

$v=\sqrt{2as}=\sqrt{2\mu gs}=\sqrt{2(0.25)(9.8)(52.5)}=16.0 m/s$

7 0

3 years ago

Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)

Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force, $(FS_p)s=-k\Delta s-q(\Delta s)^3$

$k=350N/m$

$q=750 N/m^3$

We have to find the work must you do to compress this spring 15 cm.

$\Delta s=15 cm=0.15 m$

Using 1 m=100 cm

Work done= $\int_{0}^{0.15}-Fd(\Delta s)$

W= $-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)$

$W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}$

$W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750$

$W=4.033 J$

Ideal spring work= $0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J$

Percentage increase in work= $\frac{4.033-3.938}{3.928}\times 100=2.4$ %

6 0

4 years ago

Cold air masses tend to originate from: Question 3 options: tropical areas an area where it's wintertime an area where it's summ

suter [353]

Answer:

Polar areas

Explanation:

From what I know Arctic, Antarctic, and polar air masses are cold. Polar (cold), Arctic (very cold), Equatorial (warm and very moist), and Tropical (warm).

With the options you gave, you can immeditaly elimate 1. tropical areas and 2. an area where its summertime.

Your then left with two options 1. an area where its wintertime and 2. polar areas

I then concluded that cold air masses tend to originate from POLAR AREAS.

6 0

2 years ago

Read 2 more answers