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svp [43]
3 years ago
14

Location A receives

Physics
1 answer:
Vlad1618 [11]3 years ago
7 0

Location A receives more rainfall than Location B due to the rain shadow effect.

<u>Explanation</u>:

  • Rain shadow effect is caused due to the presence of mountains.
  • A rain shadow area is an area of land that has been forced to become dry, devoid of any vegetation growth due to the blockage of precipitation by mountains. These rain shadow areas will have a dry climate.
  • The other side of the mountain would receive plenty of precipitation and therefore would be flourished with plant growth. These areas will have a cool and wet climate.
  • In this case, Location A is on the other side of the mountain and so receives more rainfall or precipitation. Meanwhile, Location B is on the rain shadow region and so receives less rainfall.
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Round to three significant figures.<br> 0.0785584 rounds to
Elan Coil [88]

Answer:

0.0786

Explanation:

zero after the decimal place is not a significant figure since it comes before the real integer "7".

"5 " in ten thousandth place is rounded off to "6" because the next digit is also another "5",

so we get the three sfg 0.0786

8 0
3 years ago
The 10-lb block has a speed of 4 ft/s when the force of f=(8t2)f=(8t2) lb is applied. determine the velocity of the block when t
KatRina [158]

The velocity of the block when t == 2 s is 60.7 ft./sec.

Equations of Motion.

Here the friction is F_f = \mu_k N = 0.2 N

+ \uparrow \sum F_y = ma_y; \quad N – 10 = \frac { 10 } { 32.2 }(0) \quad N = 10 lb \\ \begin{aligned} \underrightarrow{ + } \sum F_x = ma_x; \quad 8t^2 – 0.2(10 &) = \frac { 10 } { 32.2 }a \\ & a = 3.22(8t^2 – 2) ft/s^2 \end{aligned}

Kinematics.

The velocity of the block as a function of t can be determined by

integrating dv = adt using the initial condition v = 4 ft./s at t = 0.

\int_{ 4 ft/s }^{ v } dv = \int_0^t 3.22(8t^2 – 2)dt \\ \begin{aligned} v – &4 = 3.22 (\frac 8 3 t^3 – 2t) \\ & v = \{8.5867t^3 – 6.44t + 4 \} ft/s \end{aligned}

The displacement as a function of t can be determined by integrating

ds = vdt using

the initial condition s = 0 at t = 0

\int_0^s ds = \int_0^t (8.5867t^3 – 6.44t + 4)dt \\ s = \{2.1467t^4 – 3.22t^2 + 4t \} ft

at t = 2 sec

s = 30 ft.

Thus, at s = 30 ft.,

\begin{aligned} v &= 8.5867(2.0089^3) – 6.44(2.0089) + 4 \\ &= 60.67 ft/s \\ &= 60.7 ft/s \end{aligned}

Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.

Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system.

Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics.

Learn more about kinematics here : brainly.com/question/24486060

#SPJ4

5 0
1 year ago
Carlos uses a rope to pull his car 30 m to a parking lot because it ran out of gas. If Carlos exerts 2,000 N of force to pull th
Fiesta28 [93]
<span>5.8 × 104 J
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6 0
3 years ago
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What happens to a path of a light ray parallel to the principal axis, after it passes through a converging
Simora [160]

Answer: The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. ... All three rays should intersect at exactly the same point.

Explanation: Once these incident rays strike the lens, refract them according to the three rules of refraction for converging lenses.

3 0
3 years ago
A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti
vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

q(t) = e^{-\frac{t}{(R*C)}}

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}

Ln(\frac{1}{2}) = -\frac{0.01}{RC}

- RC = \frac{0.01}{Ln(1/2)}

RC = 0.014s

We know that RC is equal to the time constant, then

T = RC  = 0.014s = 14ms

Therefore the time constant for the process is about 14ms

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