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gtnhenbr [62]
3 years ago
14

How is a front defined in terms of weather ?. A) the leading edge of an anticyclone flow. B) the trailing edge of a warm front.

C) the physical boundary between two or more air gasses. D) the stationary boundary between a warm and cold air mass
Physics
2 answers:
Svetllana [295]3 years ago
6 0
A front is defined as a "boundary" that separates two masses of different air densities. I am not entirely sure, but I think the answers are either C or D, but I'm mostly convinced it's option D.
sergeinik [125]3 years ago
3 0

Answer: C) The physical boundary between two or more air gasses.

Explanation:

A weather front is a transitional zone between the two air masses. It forms a physical boundary between the two air masses that exhibit differences in the temperature, humidity and densities. It is responsible for bringing the change in the climatic condition of the region. The types of weather fronts are cold, warm, occluded and stationary.

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Maria designs a test to see if lemon trees that receive more water produce larger lemons.
denis-greek [22]
Do you know the answer

7 0
3 years ago
A 1.0-m-tall vertical tube is filled with 20°C water. A tuning fork vibrating at 580 Hz is held just over the top of the tube as
Mademuasel [1]

Answer:

water heights of the tube are 0.851 m  , 0.553 m, 0.255 m

Explanation:

given data

frequency = 580 Hz

temperature = 20°C

tube = 1 m

to find out

water heights of the tube

solution

we will apply here formula for length that is

length L = v ( 2n -1 ) / 4f

here v is velocity o sound that is 343.2 m/s

so for n = 1

L = 343.2 ( 2(1) -1 ) / 4(580) = 0.147931 m

for n = 2

L = 343.2 ( 2(2) -1 ) / 4(580) = 0.443793 m

for n = 3

L = 343.2 ( 2(3) -1 ) / 4(580) = 0.739655 m

for n = 4

L = 343.2 ( 2(4) -1 ) / 4(580) = 1.035517 m is greater than 1

and so here  height is measured less than 1 m

so water heights of the tube are 1 m - 0.147931 m  , 1 m - 0.443793 m, 1 m - 0.739655 m

so water heights of the tube are 0.851 m  , 0.553 m, 0.255 m

3 0
3 years ago
ILL GIVE THE BEST ANSWER BRAINLIEST I PROMISE AND ILL THANK EVERY ANSWER A polar bear is able to walk on ice without breaking it
krek1111 [17]

Answer:

Polar bears (Ursus maritimus) are of special interest because of their large size, white color and position as the top-level carnivore in the remote arctic environment. They occur only in the northern hemisphere nearly always in association with sea ice. They have only two colors of fur: tan and white. Polar bears were created to withstand cold temperatures and are quite adaptable. Polar bears are ferocious and the most dangerous of bears. They can lop a person's head off with one swoop of their paw.

5 0
3 years ago
Read 2 more answers
A 2-kg cart, traveling on a horizontal air track with a speed of 3m/s, collides with a stationary 4-kg cart. The carts stick tog
ladessa [460]

Answer:

The impulse exerted by one cart on the other has a magnitude of 4 N.s.

Explanation:

Given;

mass of the first cart, m₁ = 2 kg

initial speed of the first car, u₁ = 3 m/s

mass of the second cart, m₂ = 4 kg

initial speed of the second cart, u₂ = 0

Let the final speed of both carts = v, since they stick together after collision.

Apply the principle of conservation of momentum to determine v

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 3 + 0 = v(2 + 4)

6 = 6v

v = 1 m/s

Impulse is given by;

I = ft = mΔv = m(

The impulse exerted by the first cart on the second cart is given;

I = 2 (3 -1 )

I = 4 N.s

The impulse exerted by the second cart on the first cart is given;

I = 4(0-1)

I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).

Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.

8 0
4 years ago
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2
dybincka [34]

Complete question is;

Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.

Block 1 is removed without disturbing block 2.

Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is mass of block 3

Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a

Making T the subject gives us;

T = (m2)g - (m2)a - - - (eq 2)

where;

g is acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is mass of block 2

To solve for the acceleration, we will just substitute (m3)a for T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

3 0
3 years ago
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