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3 years ago

Which kind of air front forms when a warm air mass meets the area of a cooler air mass?

Physics

1 answer:

mihalych1998 [28]3 years ago

8 0

Answer: "a cold front" .
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A coiled telephone cord forms a spiral with 90.0 turns, a diameter of 1.30 cm, and an unstretched length of 57.0 cm. Determine t

Vinvika [58]

Answer:

2.36 μ H

Explanation:

Given,

Number of turns= 90

diameter = 1.3 cm = 0.013 m

unscratched length = 57 cm = 0.57 m

Area, A = π r²

= π x 0.0065² = 1.32 x 10⁻⁴ m²

we know,

$L = \dfrac{\mu_0N^2A}{l}$

$L = \dfrac{4\pi \times 10^{-7}\times 90^2\times 1.32\times 10^{-4}}{0.57}$

L = 2.36 μ H

Hence, the inductance of the unstretched cord is equal to 2.36 μ H

8 0

3 years ago

How is a well different from a spring?

viktelen [127]

Answer:

well people find a hole on the ground and they can see water and that theu build the well around them but the answer is. C

4 0

4 years ago

Read 2 more answers

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

$\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2$

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

$I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2$

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

$T = I\alpha = 0.303*0.6454 = 0.196 Nm$

So the force acting on the other end to generate this torque mush be:

$F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N$

4 0

4 years ago

How to solve part (b)?

maria [59]

Answer:

1.11m

Explanation:

From the diagram we are given the following forces;

F1 = 24.3N

F3 = 30N

Since the sum of upward forces is equal to that of downward force, then;

F2 = F1 + F3

F2 = 24.3N + 30N

F2 = 54.3N

Required

Distance between B and C

First we need to get Length of AC

Take moment about A

Anticlockwise moment = F3 cos20 * AC

Anticlockwise moment = 30ACcos 20

Clockwise moment = 1.2 * F2

Clockwise moment = 1.2(54.3) = 65.16Nm

Applying the principle of moment;

Sum of ACW moment = Sum of CW moments

30ACcos 20 = 65.16

AC = 65.16/30cos20

AC = 65.16/28.19

AC = 2.31m

Get the distance BC

AC = AB + BC

BC = AC-AB

BC = 2.31 - 1.2

BC = 1.11m

Hence the separation between B and C is 1.11m

Note that the force F1 got in (a) was the value used in the calculation.

8 0

3 years ago

I was having trouble with this problem, and problems like it: A 3.2 kg pelican, with a 1.73 kg fish in its mouth, is flying 1.52

Oduvanchick [21]

Answer:

28.1 m/s

Explanation:

$u_x$ = Initial velocity of the fish = 1.52 m/s

y = Height of the bird = 40 m

$a_y$ = Acceleration in y axis = $9.81\ \text{m/s}^2$

$u_y$ = Initial velocity in y axis = 0

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 40=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{40\times 2}{9.81}}\\\Rightarrow t=2.86\ \text{s}$

$v_y=u_y+a_yt\\\Rightarrow v_y=0+9.81\times 2.86\\\Rightarrow v_y=28.057\ \text{m/s}$

The final velocity in x direction will remain the same as the initial velocity as there is no acceleration in the x direction $u_x=v_x=1.52\ \text{m/s}$

Resultant velocity is given by

$v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{1.52^2+28.057^2}\\\Rightarrow v=28.1\ \text{m/s}$

The fish is moving at a velocity of 28.1 m/s when it hits the water.

6 0

3 years ago