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3 years ago

6

Two waves equal amplitudes meet other result in a wave with zero amplitudes. Which phenomenon

1 answer:

Paladinen [302]3 years ago

8 0

The wave phenomenon known as "destructive interference".

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Can you help me understand force=mass x accleration

gavmur [86]

Force = mass * acceleration

Here is an equation for force, used to calculate force.

Look at it this way:-

Joules is used to measure force, while mass is measured by kg. Acceleration is measured by m/s/s (meters per second per second, or meters per second squared.)

joules = m/s/s * kg

Which makes everything fit in just right! :D

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3 years ago

The variable ______________ describes how quickly something moves.

IrinaK [193]

it's up in Gogle trust me

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3 years ago

You are driving home from school steadily at for 180 km. It then begins to rain and you slow to You arrive home after driving 4.

nata0808 [166]

Answer:

Explanation:

Question is incomplete

Assuming the question you have asked is

You are driving home from school steadily at 95 km/h for 180 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 4.5 h.

given,

speed of 95 km/h for 180 km

due to rain

speed is reduced to 65 km/h

distance traveled in 4.5 hour

time taken to travel 180 km

d = s x t

$t = \dfrac{180}{95}$

t = 1.9 hr

distance traveled in time, t' = 4.5-1.9 = 2.6 hr

Speed of vehicle = 65 Km/h

d' = s x t'

d' = 65 x 2.6

d'= 169 Km

total distance your hometown from school

D = d + d'

D = 180 + 169

D = 349 Km

6 0

3 years ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

3 years ago

What is kelvin scale?

Darina [25.2K]

<span>a scale of temperature with absolute zero as zero, and the triple point of water as exactly 273.16 degrees.</span>

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3 years ago

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