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dezoksy [38]
3 years ago
15

On a weather map, lines joining places with the same air pressure are called ___?

Physics
2 answers:
Svetlanka [38]3 years ago
6 0

Answer:

They are Isobars

Damm [24]3 years ago
4 0

Answer:

Isobars are lines on a weather map joining together places of equal atmospheric pressure. On the map the isobar marked 1004 represents an area of high pressure, while the isobar marked 976 represents an area of low pressure. Isobars on a weather map The numbers measure the atmospheric pressure in millibars.

Explanation:

Answer from: www.bbc.co.uk/bitesize/guides/z8yp7hv/revision/2

This is the exact answer so I advise that you do not copy and paste it or you might get in trouble.

Hope it helps bai!

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A 0.35 m2 coil with 50 turns rotates at 5 radians per sec2 in a magnetic field of 0.6 Tesla. What is the value of the rms curren
soldier1979 [14.2K]

Answer:

11.25 amps

Explanation:

For transformers, the magnetic flux

\Phi _{max} = \beta  \times A

Therefore;

\Phi _{max} = 0.6  \times 0.35 = 0.21 \ Weber

Ф = Фmax (cosωt) = 0.21·(cos(5·t))

From Faraday's law of induction, we have;

ε = -N × dΦ/dt

Which gives;

dΦ/dt = -1.05(sin (5t) )

ε = -N × dΦ/dt = -50× -1.05(sin (5t) )

ε = 52.5(sin (5t) )

I = ε/R = 52.5(sin (5t) )/3.3 = 15.9091(sin (5t) ) amps

The peak current is therefore = 15.9091 amps

The rms current = Peak current /√2 = 15.9091/(√2) = 11.25 amps.

5 0
3 years ago
Someone can help me pls <br><br> Is physical science class
pentagon [3]

Answer:

<em> I can't see the picture</em>

Explanation:

6 0
3 years ago
Which steps are important when designing and conducting a scientific experiment
artcher [175]
Having your space clean. have on close toed shoes. have your hair pulled back into a ponytail. keep ur work space clean. wear gloves and goggles. do not have on droopy clothes. follow the steps on the board and double check them. 
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What dating technique could scientists use to find the age of a meteorite?
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8 0
3 years ago
An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1
motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment M_{l}=2.90\times10^{-28}\ kg

Mass of the heavier fragment M_{h}=1.62\times10^{-27}\ Kg

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c

\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2

\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}

\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})

\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}

v_{2}^2=\dfrac{c^2}{8.93}

v_{2}=0.335c

Hence, The speed of the heavier fragment is 0.335c.

7 0
3 years ago
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