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3 years ago

How can being near a large body of water affect the climate of the region?

Physics

2 answers:

shepuryov [24]3 years ago

7 0

Answer:

yes

Explanation:

If you've ever heard of lake effect snow that is when cold air mass moves across long expanses of warmer lake water. The lower layer of air, heated up by the lake water, picks up water vapor from the lake and rises up through the colder air above; the vapor then freezes and is deposited on the leeward (downwind) shore. this is most common with lake Michigan when cold air travels down from Canada making the neighboring states colder, so YES, being near any body of water can effect the climate.

I hope this helps

bija089 [108]3 years ago

7 0

Answer:

Explanation:

Hurricane winds whip up large

waves in the ocean. These waves cause

a bulge of water in the ocean known

as a storm surge (stawrm surj). As

a storm moves over a coast, the

storm surge can cause water levels to

suddenly rise, or surge, several meters.

The main dangers in a hurricane

are flying objects, strong winds, and

flooding. If possible, leave areas

threatened by a hurricane. Take

prescription medicines with you.

If you stay in a building, board up

windows and stay away from windows

and doors. Store food, bottled water,

flashlights, and battery operated

radios. Stay away from beaches

and areas that may become flooded

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A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as

Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

$T= 3600\ sec$

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=\dfrac{2\pi}{T}$

Put the value into the formula

$\omega=\dfrac{2\pi}{3600}$

$\omega=0.001745\ rad/s$

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

$a=r\omega^2$

Put the value into the formula

$a=0.55\times(0.001745)^2$

$a=1.675\times10^{-6}\ m/s^2$

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock $1.675\times10^{-6}\ m/s^2$ .

3 0

2 years ago

A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f

wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

$t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085$

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m

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2 years ago

Olympic gold medalist Michael Johnson runs one time around the track 400 meters in 38 seconds what is his displacement what is h

Sveta_85 [38]

Displacement = 0, assuming that he runs back to original position
Average velocity is displacement/ time, since displacement =0, velocity is also 0

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3 years ago

In the four-stroke internal combustion engine, fuel is ignited during the ____ stroke.

forsale [732]

The four strokes in order are the intake stroke, the compression stroke, the power stroke, and the exhaust stroke. Fuel is ignited during the power stroke.

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3 years ago

Read 2 more answers

One beam of electrons moves at right angles to a magnetic field. the force on these electrons is 4.9 x 10-14 newtons. a second b

diamong [38]

Below are the choices that can be found elsewhere:

A. (4.9 × 10-14 newtons) · tan(30°)
B. (4.9 × 10-14 newtons) · sin(30°) 
C. (4.9 × 10-14 newtons) · cos(30°) 
D. (4.9 × 10-14 newtons) · arctan(30°) 
E. (4.9 × 10-14 newtons) · arccos(30°)

Force is proportional to the angle made by the velocity with respect to the magnetic field. It is maximum when velocity is perpendicular to the magnetic field and minimum when the velocity is parallel to the magnetic field. It is proportional to sin of the angle. In this problem it will be proportional to sin(30)

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3 years ago