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3 years ago

Which type of interference occurs when two waves exactly cancel out? NEED AN ANSWER ASAP

1 answer:

4 0

If the two waves combine to produce ANY wave that smaller
than either of the originals, that's destructive interference.

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In a tug of war, two teams exerts a force of 30 N each. What is the net force acting on the rope?

Arte-miy333 [17]

The force exerted by each team is 30 N. The forces are in the direction opposite to each other.

Force exerted by team 1, $F_{1}=30 N$

Force exerted by team 2, $F_{2}=-30 N$

Net force, $F=F_{1} +F_{2}$

$=30+(-30)=0$

Therefore, net force on the rope is 0 N.

4 0

3 years ago

A simple ideal Brayton cycle uses argon as the working fluid. At the beginning of the compression, P1 = 15 psia and T1 = 70°F, t

Shtirlitz [24]

B I thing hope this helps

7 0

4 years ago

A 5.0 mm diameter proton beam carries a total current of 1.5mA. The current density in the proton beam, which increases with dis

slavikrds [6]

Answer:

Jedge = 0.076A/cm^2

Explanation:

If you want to know what is Jedge is convenient to use an integral. The total density current is given by:

$J_T=\int_{0}^{R}J_{edge}(\frac{r}{R})dr=J_{edge}(\frac{1}{R})(\frac{R^2}{2})=\frac{J_{edge}R}{2}$ (1)

But also, we have that the total current is

$I_T=J_TA\\\\J_T=\frac{I_T}{A}=\frac{1.5*10^{-3}A}{\pi(5*10^{-3})^2}=19.09\frac{A}{m^2}$

where we have used that 5mm=5*10^{-3}m.

By replacing (1) we obtain:

$J_{edge}=\frac{2J_T}{R}=\frac{2(19.09A/m^2)}{5*10^{-3}m}=7639.4\frac{A}{m^2}=0.076\frac{A}{cm^2}$

hope this helps!!

3 0

4 years ago

Read 2 more answers

An object has a mechanical energy of 1575 J and a potential energy of 1265 J.

laila [671]

A. 1575 - 1265 = 310J

B. KE 1/2 MV^2

V=√2·KE/M = √2(310)/12 V = 7.2mls

C. PE = 1265 = mgh

h= 1265/mg = 1265/(12)(92) h= 10.8m

A. KE = 1/2 mv^2 0.5(5)(12)^2 KE = 360J

B. PE = mgh = (5)(9.8)(2.6) PE = 127.4J

C. ME = KE + PE = 360 + 127.4 ME = 487.4J

5 0

4 years ago

A teacher sets up the following demonstration in which he places a clear glass rod with an index of refraction of 1.47 into a be

vlabodo [156]

Answer:

salt water

total internal reflection

Explanation:

Let's analyze this experiment, for the rod to disappear it must have happened that the light coming from it does not reach us that we are seeing from the air n = 1, therefore the light from the rod must be reflected, for this to happen we use the refractive ratio

n<u>₁</u> sin θ₁ = n₂ sin θ₂

where n₁ is the refractive index of the rod, the greatest angle that could exist in the liquid is 90º, therefore sin 90 = 1

n₁ sin θ₁ = n₂

sin θ₁ = n₂ /n₁

the only way this expression is correct is that

n₂ <n₁

that is

n₂ < 1.47

When reviewing the given materials, the only one that meets this condition is salt water by the total internal reflection

7 0

3 years ago