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3 years ago

14

Determine the force between two long parallel wires, that are separated by a dis- tance of 0.1m the wire on the left has a curre

nt of 10A and the one on the right BT has a current of 15A, both going up. 31十131

1 answer:

valina [46]3 years ago

3 0

Answer:

3 x 10^-4 N/m

Attractive

Explanation:

r = 0.1 m, i1 = 10 A, i2 = 15 A

The force per unit length between two parallel wires is given by

$F = \frac{\mu _{0}}{4\pi }\times \frac{2i_{1}i_{2}}{r}$

$F = \frac{10^{-7}\times 2\times 10\times 15}{0.1}$

F = 3 x 10^-4 N/m

Thus, the force per unit length between two wires is 3 x 10^-4 N/m, the force is attractive in nature because the direction of flow of current in both the wires is same.

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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid.

kolbaska11 [484]

Answer:

The terminal velocity of the diver is 115 m/s = 414 km/hr

Explanation:

At terminal velocity,

Fnet = mg - Fd = 0

Drag force, Fd = cρAv²/2

mg = cρAv²/2

Terminal Velocity of a body falling through a fluid as in a diver falling through air is given by

v = √(2mg/ρcA)

where m = mass of body falling through fluid = 80 kg

g = acceleration due to gravity = 9.8 m/s²

ρ = density fluid, density of air, as obtained from literature = 1.21 kg/m³

c = coefficient of drag friction of diver falling through air, as obtained from literature = 0.7

A = the area of the diver facing the fluid = 0.14 m²

v = √(2mg/ρcA) = √((2 × 80 × 9.8)/(1.21 × 0.7 × 0.14)) = 115 m/s = 115 × (3600/1000) km/hr = 414 km/hr

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3 years ago

What acceleration results from exerting a 125N force on a 0.65kg ball?

labwork [276]

first you do your pyramid f is on top and ma is
on bottom were m=mass and a=acceleration
were in this case you do f÷m=force÷mass so again in this case 125÷0.65=192.3 to be more accurate 192.3076923077

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4 years ago

What scientific device fascinated Newton as a Boy?

Wittaler [7]

Answer:

As a boy, Sir Isaac Newton was fascinated by a Windmill.

A drawing of a Windmill, which many believe to have been sketched by the young Newton over three an half centuries ago was discovered in his childhood home on a wall.

Cheers!

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4 years ago

Calcium oxide write this in a formula

kykrilka [37]

The formula for calcium oxide is CaO.

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3 years ago

An airplane is moving at 350 km/hr. If a bomb is

bogdanovich [222]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final height

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb's initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity </h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

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3 years ago