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3 years ago

Someone please help me for a brainliest but it has to be correct

Physics

2 answers:

Katarina [22]3 years ago

7 0

The answer is 34 you have to add the numbers and divide them by how many numbers there are

Maslowich3 years ago

4 0

Alright so to find the mean we have to add all the numbers (data) together and then divide it by the amount of data. So two people are 30,two at 32, one at 34 and three at 36

so our problem is....
30+30+32+32+34+36+36+36= 266

Now we do 266/the number of data used, which is 8
so
266/8=33.25

Hope that helped ;)

If you dont understand something i did ask and i can re explain

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Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.

dezoksy [38]

The De Broglie wavelength of the electron is
$\lambda=34.0 nm=34 \cdot 10^{-9} m$
And we can use De Broglie's relationship to find its momentum:
$p= \frac{h}{\lambda}= \frac{6.6 \cdot 10^{-34} Js}{34 \cdot 10^{-9} m}=1.94 \cdot 10^{-26} kg m/s$

Given $p=mv$ , with m being the electron mass and v its velocity, we can find the electron's velocity:
$v= \frac{p}{m}= \frac{1.94 \cdot 10^{-26} kgm/s}{9.1 \cdot 10^{-31} kg}= 2.13 \cdot 10^4 m/s$

This velocity is quite small compared to the speed of light, so the electron is non-relativistic and we can find its kinetic energy by using the non-relativistic formula:
$K= \frac{1}{2}mv^2= \frac{1}{2}(9.1 \cdot 10^{-31} kg)(2.13 \cdot 10^4 m/s)^2=2.06 \cdot 10^{-22} J$

3 0

3 years ago

which principles of training refers o placing increased demands on the body? A. Specificity B. cross-training c.type D.overload

Anuta_ua [19.1K]

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5 0

3 years ago

At t=0, a train approaching a station begins decelerating from a speed of 80 mi/hr according to the acceleration function a(t)=−

vredina [299]

Answer:

a) Δx = 49.23 mi , b) Δx = 5.77 mi

Explanation:

As we have an acceleration function we must use the definition of kinematics

a = dv / dt

∫dv = ∫ a dt

we integrate and evaluators

v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt

We change variables

1+ 8t = u

8 dt = du

v - v₀ = -1280 ∫ u⁻³ du / 8

v -v₀ = -1280 / 8 (-u⁻²/2)

v - v₀ = 80 (1+ 8t)⁻²

We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v

v- 80 = 80 [(1 + 8t)⁻² - 1]

v = 80 (1+ 8t)⁻²

We repeat the process for defining speed is

v = dx / dt

dx = vdt

x-x₀ = 80 ∫ (1-8t)⁻² dt

x-x₀ = 80 ∫ u⁻² dt / 8

x-x₀ = 80 (-1 / u)

x-x₀ = -80 (1 / (1 + 8t))

We evaluate for t = 0 and x₀ and the upper point t and x

x -x₀ = -80 [1 / (1 + 8t) - 1]

We already have the function of time displacement

a) let's calculate the position at the two points and be

t = 0 h

x = x₀

t = 0.2 h

x-x₀ = -80 [1 / (1 +8 02) -1]

x-x₀ = 49.23

displacement is

Δx = x (0.2) - x (0)

Δx = 49.23 mi

b) in the interval t = 0.2 h at t = 0.4 h

t = 0.4h

x- x₀ = -80 [1 / (1+ 8 0.4) -1]

x-x₀ = 55 mi

Δx = x (0.4) - x (0.2)

Δx = 55 - 49.23

Δx = 5.77 mi

3 0

3 years ago

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Hoochie [10]

I don't think the shaping of the beach or hill would be considered
weathering, especially since it says "as smaller particles are moved
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gather and make a shape.

'C' is really talking about weathering, where rocks are broken up into
stones, and the stones into smaller pieces that can be blown away
on the wind. THAT's weathering.

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charle [14.2K]

Answer: "a stationary front" .
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