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3 years ago

If the resistance of a circuit is 3 ohms, and the voltage produced by the cell in the

Physics

1 answer:

skad [1K]3 years ago

5 0

Answer:

The magnitude of the current is 4A (amperes).

Explanation:

We can use $V=IR$ , where V stands for Voltage, I stands for current, and R stands for resistance to solve this problem.

A such:

$V=12\\I=\text{ ?}\\R=3\\\\\text{Solve for I}:\\12=3I\\3I=12\\I=\frac{12}{3}\\I=4\text{A}$

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A bowling ball is dropped from rest. what is the ball’s velocity after falling for 4 seconds?

muminat

Answer:

it would be 39.2 m/s

Explanation:

After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on.

4 0

2 years ago

Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par

Murrr4er [49]

Answer:

X₃₁ = 0.58 m and X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

F₁₃ - F₂₃ = 0

F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

F₁₃ = k q₁ q₃ / r₁₃²

F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

r₁₃ = x₃- 0

r₂₃ = 2 –x₃

We substitute

k q q / x₃² = k 4q q / (2-x₃)²

q² (2 - x₃)² = 4 q² x₃²

4- 4x₃ + x₃² = 4 x₃²

5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5

x₃ = [-4 ± 9.80] 10

X₃₁ = 0.58 m

X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

7 0

3 years ago

A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th

melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore, the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F = 624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0

3 years ago

a scale model of the solar system where 50 cm represents 1.0x10 to the fifth km is actual distance what would be the dimension o

Fofino [41]

The distance between Mars and the Sun in the scale model would be 1140 m

Explanation:

In this scale model, we have:

$x_1 = 50 cm$ represents an actual distance of

$d_1 = 1.0\cdot 10^5 km$

The actual distance between Mars and the Sun is 228 million km, therefore

$d_2=228\cdot 10^6 km$

On the scale model, this would corresponds to a distance of $x_2$ .

Therefore, we can write the following proportion:

$\frac{x_1}{d_1}=\frac{x_2}{d_2}$

And solving for $x_2$ , we find:

$x_2=\frac{x_1 d_2}{d_1}=\frac{(50)(228\cdot 10^6)}{1\cdot 10^5}=1.14\cdot 10^5 cm = 1140 m$

Learn more about distance:

brainly.com/question/3969582

#LearnwithBrainly

4 0

3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago