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3 years ago

A wave can be described as

1 answer:

4 0

A wave can be described as the disturbance of particles in an area. Think about it this way: particles (matter) carry energy. For all the laws of physics to work, this energy must be "traded" somehow. This happens by miniscule vibrations in the particles, which are apparent disturbances. This creates a wave, and therefore a wave is, indeed, a disturbance.<span />

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A single mass (m1 = 3.5 kg) hangs from a spring in a motionless elevator. The spring constant is k = 278 N/m. 1)What is the dist

artcher [175]

<h2>Answer:</h2>

0.126m

<h2>Explanation:</h2>

According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;

F = k x e -------------------(i)

Where;

k = the spring's constant.

From the question, the force acting on the spring is the weight(W) of the mass. i.e

F = W -----------------------(ii)

W = m x g;

where;

m = mass of the object

g = acceleration due to gravity [usually taken as 10m/s²]

<em>From equation (ii), it implies that;</em>

F = W = m x g

<em>Now substitute F = m x g into equation(i) as follows;</em>

F = k x e

m x g = k x e ------------------(iii)

<em>From the question;</em>

m = m1 = 3.5kg

k = 278N/m

<em>Substitute these values into equation (iii) as follows;</em>

3.5 x 10 = 278 x e

35 = 278e

<em>Now solve for e;</em>

e = 35/278

e = 0.126m

Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m

3 0

3 years ago

The initial speed of a body is 7.1 m/s. What is its speed after 2.23 s if it accelerates uniformly at 2.64 m/s 2 ? Answer in uni

Nana76 [90]

13.0m/s

1.2m/s

Explanation:

Given parameters:

Initial speed of the body = 7.1m/s

time taken = 2.23s

Acceleration = 2.64m/s²

Unknown:

Final speed = ?

Solution:

Acceleration is the rate of change of velocity with time.

a = $\frac{V - U}{T}$

a = acceleration

V = final speed

U = initial speed

T = time taken

Input the variables and solve for V;

2.64 = $\frac{V - 7.1}{2.23}$

V - 7.1 = 5.9 expression 1

V = 5.9 + 7.1 = 13.0m/s

Using the same parameters, the speed after a uniform deceleration of -2.64m/s², the negative sign implies deceleration;

from expression 1;

V - 7.1 = -5.9

V = -5.9 + 7.1 = 1.2m/s

learn more:

Acceleration brainly.com/question/3820012

#learnwithBrainly

5 0

3 years ago

Help please! Will give brainliest to first correct answer!

Olegator [25]

Answer:

1) Addition of a catalyst

2) To change the reaction rate of slope B to look like slope A, simply add a catalyst to speed up the rate of reaction, giving you a higher amount of products in a shorter amount of time (line A)

Explanation:

1 and 2)Two things can alter the rate of a reaction, either the addition of a catylist which will not alter the composition of the products or reactants, but will accelerate the reaction time, or an increase in temperature will also increase the rate at which a reaction will occur.

You could choose temperature also and have the same result, it's your choice both are correct, but catalyst is the easiest.

8 0

3 years ago

You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is

madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book $(m_0)=1.7 kg$

height(h)=1 m

time taken=0.64 s

$h=ut+frac{at^2}{2}$

$1=0+\frac{a(0.64)^2}{2}$

$a=4.88 m/s^2and [tex]a=\alpha r$

where $\alpha$ is angular acceleration of pulley

$4.88=\alpha \times 5.2\times 10^{-2}$

$\alpha =93.84 rad/s^2$

And Tension in Rope

$T=m(g-a)$

$T=1.7\times (9.8-4.88)$

T=8.364 N

and Tension will provide Torque

$T\times r=I\cdot \alpha$

$8.364\times 5.2\times 10^{-2}=I\times 93.84$

$I=0.463\times 10^{-2} kg-m^2$

$I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2$

Thus mass is uniformly distributed or some more towards periphery of Pulley

4 0

3 years ago

Swinging a golf club or baseball bat are examples of ______________ stretching.

kari74 [83]

Answer:

Static stretching is the answer.

Explanation:

Static stretching is the most common form that greatly improves flexibility. However, static stretches does little to contract the muscles needed to generate powerful golf swings. Dynamic stretches help improve your range of motion while reducing muscle stiffness.

6 0

3 years ago