Which of the following statements accurately describes oceanic electric fields?

ahrayia [7]

ans D. Substance B has a greater latent heat of vaporization than substance A.

8 0

3 years ago

Assume that our computer stores decimal numbers using 16 bits - 10 bits for a sign/magnitude mantissa and 6 bits for a sign/magn

madreJ [45]

Explanation:

a) 7.5= 111.1×2°= 0.1111×2^3

which can also be written as

(1/2+1/4+1/8+1/16)×8

sign of mantissa:=0

Mantissa(9 bits): 111100000

sign of exponent: 0

Exponent(5 bits): 0011

the final for this is:011110000000011

b) -20.25= -10100.01×2^0= -0.1010001×2^5

sign of mantissa: 1

Mantissa(9 bits): 101000100

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1101000100000101

c)-1/64= -.000001×2^0= -0.1×2^{-5}

sign of mantissa: 1

Mantissa(9 bits): 100000000

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1100000000100101

5 0

3 years ago

How long does it take to raise the temperature of the air in a good-sized living room (3.00m×5.00m×8.00m) by 10.0∘C? Note that t

tekilochka [14]

Answer : The time required is, 16.1 minutes.

Explanation :

First we have to calculate the amount of heat required to increase the temperature is:

$Q=mC\Delta T\\\\Q=\rho VC\Delta T$

$(m=\rho V)$

where,

Q = amount of heat required = ?

m = mass

$\rho$ = density of air = $1.20kg/m^3$

V = volume of air

C = specific heat of air = $1006J/kg^oC$

$\Delta T$ = change in temperature = $10.0^oC$

Now put all the given values in above formula, we get:

$Q=\rho VC\Delta T$

$Q=(1.20kg/m^3)\times (3.00m\times 5.00m\times 8.00m)\times (1006J/kg^oC)\times (10.0^oC)$

$Q=1.449\times 10^6J$

Now we have to calculate the time required.

Formula used :

$t=\frac{Q}{P}$

where,

t = time required = ?

Q = amount of heat required = $1.449\times 10^6J$

P = power = 1500 W

Now put all the given values in above formula, we get:

$t=\frac{1.449\times 10^6J}{1500W}$

$t=966s\times \frac{1min}{60s}=16.1min$

Thus, the time required is, 16.1 minutes.

5 0

3 years ago

A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.4 m/s, a typical raindrop speed. How fast

kenny6666 [7]

The final velocity after the collision is 8.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, if we consider the system to be isolated (=no external unbalanced forces), the total momentum of the raindrop+mosquito must be conserved before and after the collision.

If the collision is perfectly inelastic, moreover, the raindrop and the mosquito stick together and travel at the same velocity v after the collision.

Mathematically:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1$ is the mass of the first mosquito

$u_1 = 0$ is the initial velocity of the mosquito

$m_2 = 50 m_1$ is the mass of the raindrop

$u_2 = 8.4 m/s$ is the initial velocity of the raindrop

$v$ is the final combined velocity of the raindrop+mosquito

Re-arranging the equation and substituting, we find:

$m_1 u_1 + 50 m_1 u_2 = (m_1 + 50 m_1) v\\50 m_1 u_2 = 51 m_1 v\\50 u_2 = 51 v\\v=\frac{50}{51}u_2 = \frac{50}{51}(8.4)=8.2 m/s$

Learn more about momentum here:

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brainly.com/question/6573742

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#LearnwithBrainly

4 0

3 years ago

Convection currents produce the heat in the Earth’s interior.

DaniilM [7]

Answer:

False

Explanation:

Convection currents do not produce heat. In fact, convection current are a method of transfer of heat, not of production. Convection occurs when there is a fluid which is heated from bottom, from an external source of heat (such as a pot of boiling water over a flame): the bottom part of the fluid becomes warmer, and so less dense than the colder part, therefore it starts moving up, and it is replaced by the colder parts of the fluid, which go down. Later, these colder parts become warmer, so they start going up, being replaced by new colder parts, etc... in a cycle. This is known as convection current, but it requires an external source of heat, it does not produce heat by itself.

8 0

3 years ago

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