All categories

2 years ago

--Please help!--

Physics

1 answer:

enot [183]2 years ago

6 0

Answer:

C) Amplitude.

Explanation:

By definition, the amplitude of wave is defined as the distance from the rest position to the highest or lowest point of the wave.

The highest and lowest points of the wave are called crests and troughs respectively; Therefore, the amplitude of the waves is also said to be the height of the wave from its rest position to its crest—and also the height of the from the rest position it its trough.

This means Choice C (amplitude) is the correct term used to describe the height of a wave from its rest position to its crest.

<em>P.S: Choice A, the crest height, can also be the correct term, but the term ''amplitude'' is often used and is more technical. </em>

You might be interested in

two masses are separated by 1 m. suppose the masses are moved so they are 2 m apart how will the gravitational force change

Verdich [7]

The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together

5 0

2 years ago

One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p

Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

$P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}$

$kg\dfrac{m^2}{s^2}=Joule$

$P=\dfrac{kgm^2}{s^3}$

So, the answer is D. the proper replacement unit for one joule per second

8 0

2 years ago

A 200g block on a 50cm long string swings in a circle, it's frictionless and 75rpm. What is its speed and tension on string

krek1111 [17]

Angular velocity = (75x2pie)/60
      =2.5pie ras^-1
linear velocity(or speed) at end of string, v = radius x angular velocity
       v= 0.5 x 2.5pie
   v=3.93 ms^-1

tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
      F= (0.2 x 3.93^2)/0.5
      F=6.18 N
(sorry if wrong)

3 0

2 years ago

Read 2 more answers

A shopper pushes a 5.32 kg grocery cart

Juli2301 [7.4K]

Answer:

$\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

$\text { Acceleration }=\frac{\text { force }}{\text { mass }}$

Given that,

Mass = 5.32 kg

$\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}$

$x=-28.7^{\circ}$

F = 12.7N

Normal force = mg + F sinx,

“m” being the object's "mass",

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

$\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })$

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

<u>Acceleration of the cart</u>:

$\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}$

$\text { Acceleration }=\frac{58.232}{5.32}$

$\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}$

$\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

7 0

2 years ago

Read 2 more answers

Zero field spot for opposite unequal charges

xz_007 [3.2K]

Answer:

The zero field location has to be on the line running between the two point charges because that's the only place where the field vectors could point in exactly opposite directions. It can't be between the two opposite charges because there the field vectors from both charges point toward the negative charge.

7 0

2 years ago