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2 years ago

What accounts for the two precipitation peaks in mbandaka?

1 answer:

7 0

The two precipitation peaks in Mbandaka during March to April and September to November is due to the intertropical convergence zone.

Intertropical convergence zone is a narrow zone located near the equator. It is where the northern and southern air masses intersect which results to low atmospheric pressure. Due to the intertropical convergence zone’s meeting of air masses, often times the air pressure are lower which will results to colder air, or even rainfall during the period of March to April, and most especially September to November in Mbandaka.

<span>Since Mbandaka is located at the cented of Tumba-Ngiri-Maindombe area, which is named as a Wetland of International importance, there is really a bigger chance that this place experience above 60mm precipitation in a year, temperatures averaging from 23 – 26 degrees Celsius.</span>

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Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual

kotegsom [21]

Answer:

The amount of heat required is $H_t = 1.37 *10^{6} \ J$

Explanation:

From the question we are told that

The mass of water is $m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g$

The temperature of the water before drinking is $T_w = 3.8 ^oC$

The temperature of the body is $T_b = 36.6^oC$

Generally the amount of heat required to move the water from its former temperature to the body temperature is

$H= m_w * c_w * \Delta T$

Here $c_w$ is the specific heat of water with value $c_w = 4.18 J/g^oC$

$H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)$

=> $H= 7.8 *10^{4} \ J$

Generally the no of mole of sweat present mass of water is

$n = \frac{m_w}{Z_s}$

Here $Z_w$ is the molar mass of sweat with value

$Z_w = 18.015 g/mol$

=> $n = \frac{5.7 *10^2}{18.015}$

=> $n = 31.6 \ moles$

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

$H_v = n * L_v$

Here $L_v$ is the latent heat of vaporization with value $L_v = 7 *10^{3} J/mol$

=> $H_v = 31.6 * 7 *10^{3}$

=> $H_v = 1.29 *10^{6} \ J$

Generally the overall amount of heat energy required is

$H_t = H + H_v$

=> $H_t = 7.8 *10^{4} + 1.29 *10^{6}$

=> $H_t = 1.37 *10^{6} \ J$

4 0

2 years ago

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0

2 years ago

The total charge that an automobile battery can supply without being recharged is given in terms of ampere-hours. A typical 12 V

Lelechka [254]

Answer:

7.894 Hours.

Explanation:

Based on information number hours that this battery will last with give load has mathematical relation of.

$t = \frac{60Ah}{load in amperes.}$

with load 60A t = 1h, 30A t = 2h so on and forth.

two head lights draw total current of 2x3.8A = 7.6A.

putting this in above relation gives.

$t = \frac{60Ah}{7.6A}=7.894 h$ .

That is how long will it be before battery is dead.

6 0

1 year ago

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T. If the magnetic

larisa86 [58]

Answer:

The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

Explanation:

Given that,

Charge $q =8.4\times10^{-4}\ C$

Angle = 35°

Magnetic field strength $B=6.7\times10^{-3}\ T$

Magnetic force $F=3.5\times10^{-2}\ N$

We need to calculate the velocity.

The Lorentz force exerted by the magnetic field on a moving charge.

The magnetic force is defined as:

$F = qvB\sin\theta$

$v = \dfrac{F}{qB\sin\theta}$

Where,

F = Magnetic force

q = charge

B = Magnetic field strength

v = velocity

Put the value into the formula

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times\sin35^{\circ}}$

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times0.57}$

$v = 10910.36\ m/s$

$v = 1.1\times10^{4}\ m/s$

Hence, The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

4 0

2 years ago

An object undergoing simple harmonic motion has a maximum displacement of 6.2 m at t=0.0 s. if the angular frequency of oscillat

Law Incorporation [45]

Answer:

Explanation:

From above question we know that

A = 6.2 m

f = 1.6 rad/s

t = 3.5 s

x =?

We know that,

x = Acos(2pie ft)

Putting all values in above eq.

x = 6.2 x cos(2x3.142x1.6x3.5)

x = - 4.8

Displacement can never be negative so ignore - sign.

4 0

2 years ago