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3 years ago

Surface currents are caused by ___

Physics

2 answers:

sergeinik [125]3 years ago

6 0

Surface currents are caused by the created by the global wind system

alexandr402 [8]3 years ago

3 0

They are created by the global system

You might be interested in

What is the power dissipated by the parallel resistor circuit shown above?

MatroZZZ [7]

One formula for power is

P = (voltage squared)/(resistance).

Resistance of 3 equal resistors in parallel is 1/3 of the value = 100/3 ohms.

Power = (40^2)/ (100/3)

. = (1600 • 3) / (100)

. =. 48 watts .

3 0

4 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c

EleoNora [17]

Answer:

63.8 V

Explanation:

We are given that

$A_1=1.6 cm^2=1.6\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

$A_2=1.2 cm^2=1.2\times 10^{-4} m^2$

$A_3=4.4 cm^2=4.4\times 10^{-4} m^2$

$A_4=7 cm^2=7\times 10^{-4} m^2$

Potential difference,V=140 V

We know that

$R=\frac{\rho l}{A}$

According to question

$l_1=l_2=l_3=l_4=l$

In series

$R=R_1+R_2+R_3+R_4$

$R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})$

$R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})$

$R=\rho l(18284.6)$

$I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}$

Potential across 1.2 square cm= $V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V$

Hence, the voltage across the 1.2 square cm wire=63.8 V

3 0

3 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$