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3 years ago

Thunderclouds and storms come from which type of front?

Physics

2 answers:

Lena [83]3 years ago

8 0

Thunder clouds and storms are both come from the cold front.

Answer: Option A

<u>Explanation:</u>

The cold front is the zone of transition where the warm air is lifted up by the effect of the cold air. It is actually a displacement. The cold fronts are seen moving from the North West region to the southeast region.

The air behind the cold front is dry and as the name suggests are much colder. Typically, wind becomes gusty as the cold front passes.

So, temperatures fall suddenly and heavy rain occurs, sometimes with hail, lightning, and thunder. Rising warm air in the front creates tubular clouds and thunderstorms.

alekssr [168]3 years ago

3 0

Answer:

I believe it's A. Cold Front.

Hope this helps, Mark as brainiest please. :-)

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How to solve these two questions?

hammer [34]

1) See attached figure

The relationship between charge and current is:

$i = \frac{Q}{t}$

where

i is the current

Q is the charge

t is the time

Therefore, the current is the rate of change of the charge passing through a given point over time.

This means that for a graph of charge over time, the current is just equal to the slope of the graph.

For the graph in this problem:

- Between t = 0 and t = 2 s, the slope is

$\frac{50-0}{2-0}=25 C/s$

therefore the current is

i = 25 A

- Between t = 2 s and t = 6 s, the slope is

$\frac{-50-(50)}{6-2}=-25 C/s$

therefore the current is

i = -25 A

- Between t = 6 s and t = 8 s, the slope is

$\frac{0-(-50)}{8-6}=25 C/s$

therefore the current is

i = 25 A

The figure attached show these values plotted on a graph.

2) $15 \mu C$

The previous equation can be rewritten as

$Q = i t$

This equation is valid if the current is constant: if the current is not constant, then the total charge is simply equal to the area under a current vs time graph.

Here we have the current vs time graph, so we gave to find the area under it.

The area of the first triangle is:

$A_1 = \frac{1}{2}(0.001 s)(0.010 A)=5\cdot 10^{-6} C$

While the area of the second square is

$A_2 = (0.002 s - 0.001 s)(0.010 A)=1\cdot 10^{-5}C$

So, the total area (and the total charge) is

$Q=A_1 +A_2 = 5\cdot 10^{-6} + 1\cdot 10^{-5} = 1.5\cdot 10^{-5}C=1.5 \mu C$

3 0

4 years ago

How could you increase the force advantage of a lever?Select one:a. Make the effort length longer.b. Make the effort length shor

11Alexandr11 [23.1K]

The ideal mechanical advantage of a lever (IMA) is given by:

$IMA=\frac{Le}{Lr}$

Where:

Le = Effort of the arm

Lr = Resistance arm.

Therefore, we can increase the force adventage by increasing the effort arm or reducing the load arm

Answer:

a. Make the effort length longer.

4 0

1 year ago

Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water

artcher [175]

Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x

density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %

4 0

3 years ago

Calculate ideal work (in J) when a single stream of 1 mole of air is heated and expanded from 25 C and 1 bar to 100 C and 0.5 ba

NISA [10]

Answer:

-1786.5J

Explanation:

Temperature 1=T1=25°c

Temperature 2=T2=200°c

Pressure P1=1bar

Pressure P2=0.5bars

T=37°c+273=310k

Note number if moles=1

Recall work done =2.3026RTlogp2/P1

2.3026*8.314*310log(0.5/1)

-1786.5J

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3 years ago

Which planet, when viewed through a telescope, appears as a reddish ball interrupted by some permanent dark regions that change

Andrej [43]

The answer should be Mars.

4 0

3 years ago

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