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3 years ago

Which season is occurring in earths northern hemisphere when earths Southern Hemisphere is tilted toward the sun

Physics

2 answers:

SVETLANKA909090 [29]3 years ago

6 0

The answer is winter

kenny6666 [7]3 years ago

4 0

Winter Season.

When earth's Southern Hemisphere is tilted towards the sun, it means there is summer season in the Southern Hemisphere as the sun is close towards south pole. During the same period, In case of Northern Hemisphere, it is totally opposite. When the south pole is tilted towards sun, it means the Northern Hemisphere is away from sun and thus the winter season in the Northern hemisphere.

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Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago

Match the term with its description.

Mrac [35]

A=Ion

B=pH

C=Acid

D=Alkaline

6 0

3 years ago

A 0.500-kg mass suspended from a spring oscillates with a period of 1.18 s. How much mass must be added to the object to change

olga55 [171]

Answer:

The add mass = 5.465 kg

Explanation:

Note: Since the spring is the same, the length and Tension are constant.

f ∝ √(1/m)........................ Equation 1 (length and Tension are constant.)

Where f = frequency, m = mass of the spring.

But f = 1/T ..................... Equation 2

Substituting Equation 2 into equation 1.

1/T ∝ √(1/m)

Therefore,

T ∝ √(m)

Therefore,

T₁/√m₁ = k

where k = Constant of proportionality.

T₁/√m₁ = T₂/√m₂ ........................ Equation 3

making m₂ the subject of the equation

m₂ = T₂²(m₁)/T₁²........................... Equation 4

Where T₁ = initial, m₁ = initial mass, T₂ = final period, m₂ = final mass.

Given: T₁ = 1.18 s m₁ = 0.50 kg, T₂ = 2.07 s.

Substituting into equation 4

m₂ = (2.07)²(0.5)/(1.18)²

m₂ = 4.285(1.392)

m₂ = 5.965 kg.

Added mass = m₂ - m₁

Added mass = 5.965 - 0.5

Added mass = 5.465 kg.

Thus the add mass = 5.465 kg

7 0

3 years ago

(c) Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What f

ziro4ka [17]

Answer:

Force, F = 2356.19 N

Explanation:

Given that,

Diameter of the circular tip, d = 1 mm = 0.001 m

Its radius, r = 0.0005 m

Pressure created, $P=3\times 10^9\ N/m^2$

To find,

Force exerted on the nail.

Solution,

The pressure exerted by an object is equal to the force acting per unit area. Its formula is given by :

$P=\dfrac{F}{A}$

$F=P\times A$

$F=3\times 10^9\ N/m^2\times \pi (0.0005\ m)^2$

F = 2356.19 N

Therefore, the force exerted on the nail is 2356.19 N.

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3 years ago

Why does heat transfer in metal faster than it will in air

fgiga [73]

The airrrrrr the ahkakjgennl

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3 years ago