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3 years ago

1 point

1 answer:

4 0

I think it’s 150 because if u multiply 3 x 50 it’s 150 but so it’s ore simple you do 3 x5 equals 15 then add the zero from the 50 so you’ll get 150 hope I helped

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True or false

Readme [11.4K]

Answer:

true

Explanation:

I believe the answer i chose because the earths atmosphere is mostly made up of different things that causes the earth to interact with human life and also interacts with what's in the atmosphere like energy oxygen carbon dioxide and all the stuff like that i hope its right.

6 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0

4 years ago

The equation of a wave is represented by: Y = 0.005sin 2π(0.5x – 200t) where x and y are in meters and t in seconds. Find its i.

lys-0071 [83]

Answer:

See solution below

Explanation:

The standard equation of a wave is represented by: Y = Asin 2π(x/λ – ft)

A is the amplitude

λ is the wavelength

f is the frequency

t is the time

Comparing the standard with the given equation Y = 0.005sin 2π(0.5x – 200t)

i) A = 0.005m

Amplitude is 0.005m

ii) ω = 2πf

ω = 2π(200)

ω = 400π rad/s

iii) ω = 2πf

400π = 2πf

200 = f

Swap

f = 200Hertz

iv) Period = 1/f

Period = 1/200

Period = 0.005secs

v) wave number =

vi) Wavelength λ

0.5x = x/λ

0.5 = 1/λ

λ = 1/0.5

λ = 2m

vii) velocity - fλ

velocity = 200 * 2

velocity = 400m/s

7 0

3 years ago

What are the units used to measure specific heat capacity?.

Nuetrik [128]

The units used to measure specific heat capacity is Joules per kilogram per Kelvin.

<h3>What is specific heat capacity?</h3>

It is the amount of heat absorbed per kilogram of material when the temperature rises by 1 kelvin.

Specific heat capacity C is the Joules of energy in form of heat per kilogram per Kelvin temperature. The units represented by

C = ___ J/kg.K

Thus, the units used to measure specific heat capacity is Joules per kilogram per Kelvin.

Learn more about specific heat capacity.

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6 0

3 years ago

HELP ASAPP!!!! MARKING BRAINLIEST!

Morgarella [4.7K]

Answer:

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Explanation:

8 0

3 years ago

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