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3 years ago

How does the amount of precipitation affect the amount of weathering ?

1 answer:

7 0

Answer: Rainfall and temperature can affect the rate in which rocks weather. High temperatures = more rainfall increasing the rate of chemical weathering. Rocks in tropical regions exposed to a lot of rainfall and hot temperatures weather faster than similar rocks in cold, dry climates.

Explanation:

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A horizontal spring with a spring constant 100 N/m is compressed 20cm and used to launch a 2.5kg box across a frictionless, hori

Verdich [7]

Answer:

d = 0.544 m

Explanation:

To solve this problem we must work in two parts: one when the surface has no friction and the other when the surface has friction

Let's start with the part without rubbing, let's find the speed that the box reaches., For this we use the conservation of mechanical energy in two points: maximum compression and when the box is free (spring without compression)

Initial, maximum compression

Em₀ = Ke = ½ k x²

Final, free box without compressing the spring

$Em_{f}$ = K = ½ m v²

Emo = $Em_{f}$

½ k x² = ½ m v²

v = √ (k / m) x

Let's reduce the SI units measures

x = 20 cm (1m / 100cm) = 0.20 m

v = √ (100 / 2.5) 0.20

v = 1,265 m / s

Let's work the second part, where there is friction. In this part the work of the friction force is equal to the change of mechanical energy

$W_{fr}$ = ΔEm = $Em_{f}$ - Em₀

$W_{fr}$ = - fr d

Final point. Stopped box

$Em_{f}$ = 0

Starting point, starting the rough surface

Em₀ = K = ½ m v²

With Newton's second law we find the force of friction

fr = μ N

N-W = 0

N = W = mg

fr = μ mg

-μ m g d = 0 - ½ m v²

d = ½ v² / (μ g)

Let's calculate

d = ½ 1,265² / (0.15 9.8)

d = 0.544 m

6 0

3 years ago

Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

3 years ago

One property that electromagnetic waves differ from other types of waves is that they can

almond37 [142]

B.) travel in a vacuum

6 0

3 years ago

Which does a reference point provide

mr_godi [17]

Point of reference - an indicator that orients you generally; "it is used as a reference for comparing the heating and the electrical energy involved"

5 0

3 years ago

A student pushes a 40-N block across the floor for a distance of 10 m. How much work was applied to move the block? A. 4 J B. 40

damaskus [11]

Now for this problem, what is given is a 40 Newtons which would represent the force to be applied to the object, and a distance of 10 meters after the application of the said force. When these two combine, work is done. The unit for work is Joules and this is what we are looking for. The formula to get Joules or for work would be the force applied to the object multiplied by the distance that it travelled after the application of the force. It looks like this

work = force x distance
Joules = Newtons x meter

so let us substitute the variables to their corresponding places

Joules = 40 N x 10 m
Joules = 400 J

So the answer to this question would be C. 400 J

7 0

3 years ago

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