The Earth moves in predicable patterns. One of these patterns causes the seasons as

A scientist classifies some plants into two groups

Answer:

I'm pretty sure the answer is C . since deciduous trees are trees that loose leaves seasonally and coniferous trees are trees that don't loose leaves seaosnally and survive through the winter.

8 0

3 years ago

Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv

deff fn [24]

Answer:

electric field E = - k Q (1 /r(r-a)), force F = - k Q qo / r (r-a) and force for r>>a F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

λ = q / x

As it is constant, we can write it based on differentials

λ = dq / dx

dq = λ dx

We already have all the terms, let's integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

E = k ∫ λ dx / x²

E = k la (- 1 / x)

Let's get the negative sign from the parentheses

E = - k λ (1 / x)

E = - k λ (1 /(r-a) -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

E = - k Q/a [a / r (r-a)]

E = - k Q (1 /r(r-a))

b) We calculate the force.

F = E qo

F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

F = - k Qqo / r² (1 - a/r)

F ≈ - k Q qo / r²

6 0

3 years ago

A bimetallic strip (brass/steel), which is straight at room temperature, will be immersed in boiling water and allowed to equili

lakkis [162]

The thermal expansion of the materials allows to find the deflection of the bimetallist strip is Δy = 3.48 cm

given paramers

* Bimetallic brass / steel tape

* Initial temperature, room temperature T = 20ºC

* Final temperature, boiling water = 100ºC

* initial length L₀ = 222mm (1cm / 10mm) = 22.2cm

* thickness of bimetallic tape e = 0.036 inch (2.54 cm/1 inch) = 0.0914 cm

to find

* perpendicular deviation or deflection (Δy)

Thermal expansion is the phenomenon of change in the length of a body due to the change in temperature, due to the increase in the length of the atomic and molecular bonds, macroscopically it is described by

ΔL = α L₀ ΔT

ΔL and ΔT are the variation of the length and temperature respectively, L₀ is the initial length and α the coefficient of expansion ends.

In this case we have a strip formed by two materials with different coefficient of thermal expansion,

Brass α_{brass} = 19 10⁻⁶ ºC⁻¹

Steel α_{steel} = 11 10⁻⁶ ºC⁻¹

In the attached we can see a diagram of the process, as the temperature increases, the material with greater thermal expansion lengthens more, so the system must curve towards the center of the material with less

thermal expansion. Let's find the length of the strip for each material

brass L_{f brass} - L₀ = α_{brass} L₀ ΔT

Steel L_{f steel} - L₀ = \alpha_{steel} L₀ ΔT

Note that the initial length is the same for the two materials and that the strip is in thermal equilibrium at room temperature.

If we assume that we have an arc of circumference, we can write the length of the arc

θ = L / r

where θ is the angle in radines, L the length of the arc and r the radius of curvature, let's write this equation for each material

brass L_{f \ brass} =θ r₁

steel L_{f \ steel} = θ r₂

we substitute in our equations

θ r₁ - L₀ = α_{brass} L₀ ΔT

θ r₂ - L₀ = α_{steel} L₀ ΔT

let's subtract the two equations

θ (r₁- r₂) = L₀ ΔT (α_{brass} - α_{steel})

the thickness of the strip is

e = r₁ -r₂

θ = $Lo \ \Delta T \ \frac{\alpha_{brass} - \alpha_{steel}}{e}$

we calculate

θ = $22.2 \ (100-20) \ \frac{(19-11) \ 10^{-6}}{0.0914}$

θ = 0.155 rad

Let's use trigonometry to find the perpendicular deflection

tan θ = Δy / L₀

Δy = L₀ tan θ

Δy = 22.2 tan 0.155

Δy = 3.48 cm

Using the thematic expansion of the two materials we find the deflection of the bimetallist strip is 3.38 cm

Learn more about thermal expansion here: brainly.com/question/18717902

7 0

3 years ago

3

Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

$KE = \frac{1}{2}mv^2$