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BARSIC [14]
3 years ago
10

What happens in a global convection cell? apex​

Physics
1 answer:
zysi [14]3 years ago
4 0

Answer:

In a global convection cell less –dense air at the equator rises and flows towards the poles. And from pole, the dense air sinks down and flows back towards the equator.... This movement of air is also supported by the Earth's rotation known as Coriolis Effect.

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However, when is the best time to fill a balloon? Helium has a tendency to escape from any opening in the balloon and will event
alekssr [168]

Answer:The small, individual helium molecules can escape through the tiny holes in the latex far more easily than the conjoined oxygen or nitrogen molecules can. ... This is why your helium balloons deflate faster than the ones you fill with air.

Explanation:

3 0
2 years ago
a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone​
Dafna1 [17]

Answer:

The centripetal acceleration of the stone is 5 m/s²

Explanation:

The length of the string to which the stone is attached, r = 1 m

The speed with which the string is rotated, v = 5 m/s

The centripetal acceleration, a_c, is given as follows;

a_c = \dfrac{v^2}{r}

Therefore, the centripetal acceleration of the stone found as follows;

a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2

The centripetal acceleration of the stone, a_c = 5 m/s².

5 0
2 years ago
Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0
3 years ago
A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is
ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0

Solving the above equation we get

t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89

So, the time the package was in the air is 1.97 seconds

3 0
3 years ago
An electric motor rotating a workshop grinding wheel at a rate of 151 rev/min is switched off. Assume constant angular decelerat
tamaranim1 [39]

Answer:

Explanation:

Initial angular velocity ω₀ = 151 x 2π / 60

= 15.8  rad /s

final velocity = 0

Angular deceleration α = 2.23 rad / s

ω² = ω₀² -  2 α θ

0 = 15.8² - 2 x 2.23 θ

= 55.99  rad

one revolution = 2π radian

55.99 radian = 55.99  / 2 π no of terns

= 9 approx .

8 0
3 years ago
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