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3 years ago

Answer the question based on this waveform.

Physics

1 answer:

Nuetrik [128]3 years ago

6 0

Answer:

Cannot be determined from the given information

Explanation:

Given the following data;

Velocity = 24 m/s

Period = 3 seconds

To find the amplitude of the wave;

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

t is the time period measured in seconds.

ϕ is the phase angle.

Hence, the information provided in this exercise isn't sufficient to find the amplitude of the waveform.

However, the given parameters can be used to calculate the frequency and wavelength of the wave.

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A heavy ball with a weight of 100 NN is hung from the ceiling of a lecture hall on a 4.4-mm-long rope. The ball is pulled to one

allochka39001 [22]

Answer:

175.3 N

Explanation:

The motion of the ball is a uniform circular motion, therefore the net force on it must be equal to the centripetal force.

There are two forces acting on the ball at the lowest point of motion:

- The tension in the string, T , upward

- The weight of the ball, $mg$ , downward

The net force (centripetal force) has the same direction as the tension (upward, towards the centre of the circular path), so we can write:

$T-mg=m\frac{v^2}{r}$

where the term on the right is the expression for the centripetal force, and where:

T is the tension in the string

$mg=100 N$ is the weight of the ball

$m=\frac{mg}{g}=\frac{100}{9.8}=10.2 kg$ is the mass of the ball

v = 5.7 m/s is the speed of the ball at the lowest point

r = 4.4 m is the length of the rope, so the radius of the circle

Solving for T, we find the tension in the string:

$T=mg+m\frac{v^2}{r}=(100)+(10.2)\frac{5.7^2}{4.4}=175.3 N$

7 0

3 years ago

How long will it take for a radioactive isotope with a decay constant of 0.15 (which means a half life of 4.6 days ) to decay to

jek_recluse [69]

Radioactive decay is given by:
N = No x e^(-λt)
We know that N/No has to be 0.05
λ = 0.15
0.05 = e^(-0.15t)
t = ln(0.05)/(-0.15)
t = 19.97 days

6 0

3 years ago

A 2.3 m long wire weighing 0.075 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of

ZanzabumX [31]

Answer:

Explanation:

Magnetic field near current carrying wire

= $\frac{\mu_0}{4\pi} \frac{2i}{r}$

i is current , r is distance from wire

B = 10⁻⁷ x $\frac{2\times49}{r}$

force on second wire per unit length

B I L , I is current in second wire , L is length of wire

= 10⁻⁷ x $\frac{2\times49}{r}$ x 33 x 1

= 3234 x $\frac{10^{-7}}{r}$

This should balance weight of second wire per unit length

3234 x $\frac{10^{-7}}{r}$ = .075

r = $\frac{3234}{.075}$ x 10⁻⁷

= .0043 m

= .43 cm .

5 0

3 years ago

I WILL MARK U BRAINLIEST IF U ANSWER THIS QUESTION! NEED IT ASAP PLS

Anuta_ua [19.1K]

One common use of a convex mirror is as shaving mirror.
One common use of convex mirror is as rear-view mirrors in automobiles vehicles.

<h3>What is a concave mirror?</h3>

A concave mirror is also referred to as a converging mirror and it can be defined as a type of mirror that is designed and developed with a reflective surface that is typically curved inward and away from the source of light.

Basically, one common use of a convex mirror include the following:

Shaving mirrors

Searchlights

Dental mirrors.

<h3>What is a convex mirror?</h3>

A convex mirror is also referred to as a diverging mirror and it can be defined as a type of mirror that is designed and developed with a reflective surface that typically bulges outward toward the source of light.

Basically, one common use of convex mirror is as rear-view mirrors in automobiles vehicles.

Read more on convex mirror here: brainly.com/question/24175067

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2 years ago

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

Anni [7]

Answer:

The phase difference is $\Delta \phi = 180^o$

Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum