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Travka [436]
3 years ago
14

A long, straight wire carrying a current is placed along the y-axis. If the direction of the current is in the +y direction, wha

t is the direction of the magnetic field due to this wire?
Physics
1 answer:
7nadin3 [17]3 years ago
7 0

Answer:

The direction of magnetic filed will be clockwise along y axis.

Explanation:

According to right hand rule,

Thumb of right hand shows the the direction of current and folded finger shows the direction of magnetic field.

So, a long and straight wire carrying a current and the direction of current is in the + y direction then the direction of magnetic filed will be clockwise across y axis.

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If the distance between two objects is decreased to 1 10 of the original distance, how will it change the force of attraction be
aleksandr82 [10.1K]

(A) It will 100 times larger than the original force.

6 0
3 years ago
A telephone pole has three cables pulling as shown from above, with F⃗ 1=(300.0iˆ+500.0jˆ) , F⃗ 2=−200.0iˆ , and F⃗ 3=−800.0jˆ .
Ray Of Light [21]

A) Net force in component form: F=100.0i-300.0j

B) Magnitude of the net force: 316.2, direction: -71.6^{\circ}

Explanation:

A)

The three forces given in this problem are:

F_1=300i+500j

F_2=-200i

F_3=-800 j

The three forces are given in component form, where the components with unit vector i is the component along the x-direction, while the components with unit vector j is the component along the y-direction.

In order to find the net force in component form, we just need to add the components of the three forces along each direction. Therefore:

- Along the x-direction:

F_x = F_{1x}+F_{2x}+F_{3x}=300+(-200)+0=100

- Along the y-direction:

F_y=F_{1y}+F_{2y}+F_{3y}=500+0+(-800)=-300

So, the net force in component form is

F=100.0i-300.0j

B)

The magnitude of a vector F is given by Pythagorean's theorem:

|F|=\sqrt{F_x^2+F_y^2}

where in this problem,

F_x=100 is the x-component

F_y=-300 is the y-component

Substituting,

|F|=\sqrt{(100)^2+(-300)^2}=316.2

The direction instead is given by

\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-300}{100})=-71.6^{\circ}

where the negative sign means the direction is below the positive x-axis.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

8 0
3 years ago
A tiger runs one kilometer. The tiger does 15,000 J of work and has a total power output of 500 W. How many minutes does it take
defon
Power is the amount of energy consumed per unit time. Having no direction, it is a scalar quantity. <span>As is implied by the equation for </span>power<span>, a unit of </span>power <span>is equivalent to a unit of work divided by a unit of time. The formula would be as follows:

P = W/t 

We calculate as follows:

500 W = 15000 J / t
t = 30 s</span>
3 0
3 years ago
The position-time equation for a certain train is
astraxan [27]

Answer:

a=4.8m/s^2

Explanation:

Hello,

In this case, since the acceleration in terms of position is defined as its second derivative:

a=\frac{d^2x(t)}{dt^2}=\frac{d^2}{dt^2}(2.9+8.8t+2.4t^2)

The purpose here is derive x(t) twice as follows:

a=\frac{d^2x(t)}{dt^2}=\frac{d}{dt}(8.8+2*2.4*t)\\ \\a=4.8m/s^2

Thus, the acceleration turns out 4.8 meters per squared seconds.

Best regards.

8 0
4 years ago
A 50-cm-long spring is suspended from the ceiling. A 270 g mass is connected to the end and held at rest with the spring unstret
AveGali [126]

Answer:

k = 22.05 N/m

Explanation:

The potential energy of the mass is converted into potential energy of the spring.

Given:

mass m = 0.27 kg

gravitational constant g = 9.8 m/s²

distance falling/ stretching of spring h = 0.24 m

U_{gravity} = U_{spring}\\ mgh = \frac{1}{2} kh^{2}

Solving for k:

k = 2mg\frac{1}{h}

5 0
4 years ago
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