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3 years ago

Which front is a head on collision or neither warm nor cold air moves

1 answer:

3 0

Stationary wave fronts are formed when two air masses of different densities and temperature range push each other but unable to move the front in any direction.

<u>Explanation:</u>

Weather Fronts

When collision happens between air masses of two with varied densities and the temperature, they form a front. These air masses also differ in humidity. When a cold front and warm front generated and push each other equally, the net force turns to be zero and there is no movement in any of the front.

By this, both of the fronts form a stationary row which separates both the fronts. Both kind of winds flow perpendicular to each other rather than parallel and this what holds the stationary front in place.

If the winds change the directions, the front may also move in the direction of the wind which is dominating the the other one. The weather seems to be often cloudy with this front. Hence, in the state of the formation of a stationary front, neither warm nor cold air moves parallel.

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A 35.4 kg girl is riding on the back of a 15.23 kg cart. the cart and the kid are both moving eastward at 4.25 m/s when she step

Grace [21]

Answer:

The final velocity of the cart is $v_c = 7.02 \ m/s$

Explanation:

From the question we are told that

The mass of the girl is $m_g = 35.4 \ kg$

The mass of the cart is $m_c = 15.23 \ kg$

The speed of the cart and kid(girl) is $v = 4.25 \ m/s$

The final velocity of the girl is $v_g = 3.06 \ m/s$

Let assume that velocity eastward is positive and velocity westward is negative (Note that if we assume vise versa it wouldn't affect the answer )

The total momentum of the system before she steps off the back of the cart

is mathematically evaluated as

$p__{T1}} = (m_g + m_c) * v$

substituting values

$p__{T1}} = (35.4 + 15.23) * 4.25$

$p__{T1}} =215.17 \ kg m /s$

The total momentum after she steps off the back of the cart is mathematically evaluated as

$p__{T2}} = (m_g * v_g ) +( m_c * v_c )$

Where $v_c$ is the final velocity of the cart

substituting values

$p__{T2}} = (35.4 * 3.06 ) +( 15.23 * v_c )$

$p__{T2}} = 108. 324 + 15.23 v_c$

Now according to the law of conservation of momentum

$p__{T1}} =p__{T2}}$

$215.17 \ kg m /s = 108. 324 + 15.23 v_c$

=> $v_c = 7.02 \ m/s$

Since the value is positive it implies that the cart moved eastward

7 0

3 years ago

A ceiling fan is turned on and reaches an angular speed of 132 rev/min in 20 s. It is then turned off and coasts to a stop in an

Mkey [24]

Answer:

Explanation:

α = (ωf - ωi)/t

acceleration phase

ωf = 132 rev/min (2π rad/rev / 60 s/min) = 4.4π rad/s

α₁ = (4.4π - 0)/20 = 0.22π rad/s²

α₂ = (0 - 4.4π)/40 = - 0.11π rad/s²

α₁/α₂ = 0.22π/- 0.11π = -2

8 0

2 years ago

A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?

Leno4ka [110]

<h3><u>Question</u><u>:</u></h3>

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?

<h3><u>Statement:</u></h3>

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.

<h3><u>Solution</u><u>:</u></h3>

Initial velocity (u) = 70 m/s
Acceleration (a) = -14 m/s^2
Time (t) = 3 s
Let the velocity of the car after 3 s be v m/s
By using the formula,

v = u + at, we have

$v = 70 + ( - 14)(3) \\ = > v = 70 - 42 \\ = > v = 28$

So, the velocity of the car after 3 s is 28 m/s.

<h3><u>Answer:</u></h3>

The car's speed after 3 s is 28 m/s.

Hope it helps

3 0

3 years ago

The space velocity of a star is 120 km/s and its radial velocity is 72 km/s. what is its tangential velocity?

Marina CMI [18]

You divide 120 and 72 and get 1.67 repeated i think

3 0

3 years ago

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser

ELEN [110]

Answer:

a) v² = G M R³, b) T = 2π / $\sqrt{GMR}$ , c) $m \sqrt{GMR^5 }$

Explanation:

a) The kinetic energy is

K = ½ m v²

to find the velocity let's use Newton's second law

F = m a

acceleration is centripetal

a = v² / R

force is the universal force of attraction

F = G m M / r²

we substitute

G m M R² = m v² R

v² = G M R³

the kinetic energy is

K = ½ m G M R³

b) angular and linear velocity are related

v = w R

w = v / R

w = $\frac{\sqrt{GMR^3 }}{R}$

w = $\sqrt{GMR}$

the angular velocity is related to the period

w = 2π / T

T = 2π / w

we substitute

T = 2π / $\sqrt{GMR}$

c) the angular moeomto is

L = m v r

L = m RA G M R³ R

L = $m \sqrt{GMR^5 }$

4 0

3 years ago