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3 years ago

Which phenomenon occurs when one wave is superimposed on another ?

2 answers:

6 0

Answer: D
When waves interfere with each other and when they meet, they continue traveling as if they had never encountered each other.

Artyom0805 [142]3 years ago

3 0

Answer:

interference

Explanation:

a p e x :))

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A wire with a current of 3.40 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at

Sloan [31]

Answer:

0.107 m

Explanation:

The magnetic field at the center of a current-carrying loop is given by

$B=\frac{\mu_0 I}{2r}$

where

$\mu_0$ is the vacuum permeability

I is the current

r is the radius of the loop

In this problem we have

I = 3.40 A is the current in the loop

$B=20 \mu T=20\cdot 10^{-6}T$ is the magnetic field at the centre of the loop

So, solving the formula for r we find

$r=\frac{\mu_0 I}{2B}=\frac{(12.56\cdot 10^{-7} H/m)(3.40 A)}{2(20\cdot 10^{-6} T)}=0.107 m$

7 0

3 years ago

If the lens in a person's eye is too highly curved, this person is suffering from?

Xelga [282]

If the lens in a person's eye is too highly curved, this person is suffering from: nearsightedness.

lens is a piece of transparent material (such as glass) that has two opposite regular surfaces either both curved or one curved and the other plane and that is used either singly or combined in an optical instrument for forming an image by focusing rays of light.

The lens is used to form an image of an object by focusing rays of light from the object.

Various types of lenses are :

Concave lens.
Convex lens.
Plano lens.

learn more about lens here

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7 0

1 year ago

Explain velocity ratio. Explain and define the efficiency of a machine

Norma-Jean [14]

Velocity ratio is also defined as the ratio of a distance through which any part of a machine moves, to that which the driving part moves during the same time. An object has a mechanical advantage if it exerts a force higher than the velocity ratio.

8 0

3 years ago

If a force does a positive amount of work on an object, does the object's speed

Alexxandr [17]

The speed of the object increases

Explanation:

We can answer this question by applying the work-energy theorem, which states that the work done on an object is equal to the change in kinetic energy of the object. Mathematically:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

W is the work done on the object

$K_f, K_i$ are the final and initial kinetic energy of the object, respectively

m is the mass of the object

v is its final speed

u is its initial speed

In this case, the force does a positive amount of work on the object, so

$W>0$

This also implies that

$K_f > K_i$

And so

$\frac{1}{2}mv^2 > \frac{1}{2}mu^2$

And therefore

$v>u$

which means that the speed of the object increases.

Learn more about work:

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8 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago