Ask question

Ask question

All categories

3 years ago

15

Which of the following describe wind patterns created by a low-pressure front, where two air masses are interacting? Choose one

answer below.

1 answer:

viva [34]3 years ago

6 0

Answer:

I think its might Be number B

Explanation:

tell me if im rong

You might be interested in

A beaker weighs 0.4N when empty and1.4N when filled with water what does ot weigh when filled with brine of density 1.2 g/cm3

PtichkaEL [24]

Answer:2.47

Explanation:

So, the beaker weighs 1.40N when filled with water, brine of density weighs about 1.7N, you add the density + water. Have a good day!

7 0

3 years ago

Which of the following materials is attracted to a magnet?

Butoxors [25]

Answer:

A. aluminum

Explanation:

because it's metal

4 0

3 years ago

Read 2 more answers

A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting

ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

$T=\dfrac{38}{8}\\\\T=4.75\ s$

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

k = spring stiffness constant

So,

$k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m$

When the cord is in air,

mg=kx

x = the extension in the cord

$x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m$

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0

3 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter

kifflom [539]

Answer:

a). 87.5 mA or $87.5 x10^{-3}$ A

b). 1.78 $\frac{m}{s}$

Explanation:

$d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}$

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

$I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\ I=87.5 x10^{-3}$ A

b).

$I=n*abs (q)*V_{d}*A$

$A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2} \\A=5.309x10^{-6}$

$V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}$

8 0

3 years ago

A circular loop carrying a current of 1.6 A is oriented in a magnetic field of 0.30 T. The loop has an area of 0.14 m 2 and is m

konstantin123 [22]

Answer:

The torque on the loop is $2.4 \times 10^{-2}$ Nm

Explanation:

Given:

Current $I = 1.6$ A

Magnetic field $B = 0.30$ T

Area of loop $A = 0.14$ $m^{2}$

Angle between magnetic field and area vector $\theta =$ 21°

Form the formula of torque in case of magnetic field,

г $= MB \sin \theta$

Where $M =$ magnetic moment

$M = IA$

г $= IAB \sin 21$

г $= 1.6 \times 0.30 \times 0.14 \times 0.3583$

г $=2.4 \times 10^{-2}$ Nm

Therefore, the torque on the loop is $2.4 \times 10^{-2}$ Nm

7 0

3 years ago