Ask question

Ask question

All categories

3 years ago

15

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

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!

You might be interested in

Which one of the following accurately pairs the device with its function?

Westkost [7]

<span>D is the correct answer. A Bourdon gage is a popular and commonly used kind of gauge for measuring pressure and vacuum. One use for a Bourdon gage is to indicate steam pressure.</span>

5 0

3 years ago

Read 2 more answers

The IMA of the pulley show is

Lera25 [3.4K]

The IMA of the pulley shown is 2.

4 0

3 years ago

Read 2 more answers

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

7 0

3 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

3 years ago

calculate the efficiency of a light bulb that had an input of 400 j transfers 100j as a light and 300j as heat.

beks73 [17]

Efficiency = useful energy out / total energy in x 100
= 100/400 x 100
=0.25 x 100
= 25%

25%

6 0

2 years ago