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4 years ago

14

What role do currents play in transporting heat? How does water temperature affect land nearby?

1 answer:

yarga [219]4 years ago

7 0

When warm currents move over cool land surface, they dissipate heat due to the temperature gradient . They hence warm up the land they blow over. If a cold current blows over warm land, the land loses heat to the current. This is how currents distribute heat across the earth.

When the current blow from water to land, the current will be warm or cold depending on the water temperatures. Therefore, they will warm up or cool the adjacent land mass.

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A capacitor is charged to a potential difference of 12v it delivers 40% of its stored energy to a lamp what is the final potenti

Korolek [52]

Call the capacitance C.
<span>Note the energy in a capacitor with voltage V is E =½CV². </span>

<span>Initial energy = ½C(12)² = 72C </span>

<span>40% of energy is delivered, so 60% remains.in the capacitor. </span>
<span>Remaining energy = (60/100) x 72C =43.2C </span>

<span>If the final potential difference is X, the energy stored is ½CX² </span>
<span>½CX² = 43.2C </span>
<span>X² = 2 x 43.2 = 86.4 </span>
<span>X = 9.3V</span>

6 0

3 years ago

A traveling wave has displacement given by y(x,t)=(2.0cm)×cos(2πx−4πt), where x is measured in cm and t in s. what is the speed

quester [9]

Answer:

v = 2 cm/s

Explanation:

The equation of the wave is

y(x,t) = (2.0cm)*cos(2π*x−4π*t)

Where,

x is measured in cm

t in s

A more general formula for this equation would be

y(x,t) = A*cos(k*x−ω*t)

Where,

A = amplitude.

k = the wavenumber

ω = the angular frequency

The velocity of the wave corresponds to

v = ω/k

v = 4π / 2π = 2 cm/s

v = 2 cm/s

8 0

3 years ago

Planets beyond the solar system.On October 15, 2001, a planet was discovered orbiting around the star HD68988. Its orbital dista

Evgesh-ka [11]

Answer:

Part a)

$M = 2.31 \times 10^{30} kg$

Part b)

$M = 1.15 M_s$

Explanation:

Part a)

As we know that the period of revolution of the planet is given by

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

now we know that

$r = 10.5 \times 10^6 km = 1.05 \times 10^{10} m$

T = 6.3 days

$T = 5.44 \times 10^5 s$

now we have

$5.44 \times 10^5 = 2\pi\sqrt{\frac{(1.05\times 10^{10})^3}{(6.67 \times 10^{-11})M}}$

$M = 2.31 \times 10^{30} kg$

Part b)

As we know that mass of sun is given as

$M_s = 2.0 \times 10^{30} kg$

now we have

$\frac{M}{M_s} = \frac{2.31 \times 10^{30}}{2 \times 10^{30}}$

$M = 1.15 M_s$

6 0

4 years ago

Review Conceptual Example 7 before starting this problem. A uniform plank of length 5.0 m and weight 225 N rests horizontally on

Andreas93 [3]

Answer:

0.82 m

Explanation:

We are given that

Weight, $w_1$ =225 N

Length,l=5 m

d=1.1 m

We have to find the distance x for which person weighs 385 N walk on the overhanging part of the plank before it just begins to tip.

Half length= $\frac{l}{2}=\frac{5}{2}=2.5 m$

$r=\frac{l}{2}-d=2.5-1.1=1.4 m$

$w_2=385 N$

When system in equilibrium then

$w_1r=w_2 x$

$225\times 1.4=385\times x$

$x=\frac{225\times 1.4}{385}$

$x=0.82 m$

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Helga [31]

Answer:

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Explanation:

5 0

3 years ago

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