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serg [7]
3 years ago
11

The average weather of a particular place is

Physics
2 answers:
qaws [65]3 years ago
8 0
Climate
prevailing over a long period
aniked [119]3 years ago
3 0
<span>The average weather of a particular place is "Climate"

In short, Your Answer would be Option B

Hope this helps!</span>
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I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app
Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

  • initial velocity of the sprinter, u = 18 km/h
  • final velocity of the sprinter, v = 27 km/h
  • time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\

The time of motion is seconds;

t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s

The sprinter’s average acceleration is calculated as follows;

a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

6 0
2 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
You climb to the top of a ladder. You drop a a ball from height H, it reaches the ground with speed V if there is no air resista
olchik [2.2K]

Answer:

heymelissa its amanda i hate ms spearman

Explanation:

yuh

6 0
3 years ago
Explain how its possible for a compound to have both iconic and covalent bonds.<br><br>Thanks!
Evgesh-ka [11]
The atoms which make up the ion are covalently bonded to one another. 19) It is possible for a compound to possess both ionic and covalent bonding. a. If one of the ions is polyatomic then there will be covalent bonding within it.
3 0
3 years ago
What are 3 different forces that act on objects on the earth ΩvΩ
Stolb23 [73]
A force is a push or pull acting upon an object as a result of its interaction with another object. There are a variety of types of forces. a variety of force types were placed into two broad category headings on the basis of whether the force resulted from the contact or non-contact of the two interacting objects.

Contact Forces

Action-at-a-Distance Forces

Frictional Force
Gravitational Force
Tensional Force
Electrical Force
Normal Force
Magnetic Force
Air Resistance Force
Applied Force
Spring Force


These are types of individual forces



Applied Force
Gravitational Force
Normal Force
Frictional Force
Air Resistance Force
Tensional Force
Spring Force
6 0
3 years ago
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