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3 years ago

What happens when two forces act in the same direction?

1 answer:

3 0

When two forces act in the same direction, they add together. ... Equal forces acting in opposite directions are called balanced forces. Balanced forces acting on an object will not change the object's motion. When you add equal forces in opposite direction, the net force is zero.

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Calculate the force on an object that has a mass of 12kg and an acceleration of 4m/s2.

iVinArrow [24]

<span>By Newton's second law of motion, we know that the resultant force acting on a body is directly proportional to the mass of the body and directly proportional to its acceleration. In system international (SI) units, the value of the constant of proportionality constant is 1. Therefore, the equation for Newton's second law of motion becomes: F = ma, where F is the resultant force, m is the mass and a is the acceleration of the object. Substituting the values of m and a into this formula, we get the result: F = 12 x 4 = 48. The SI unit for force is the Newton; therefore, <u>the answer is 48 Newtons.</u></span>

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3 years ago

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Who is the father of electricity?

Tpy6a [65]

William Gilbert is known as the father of electricity.

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3 years ago

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Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

7 0

3 years ago

Help in physics please :(((

Arlecino [84]

Answer:

I am sorry I can't draw graphical ok how to draw the graph where what is your position the displacement of time and work 7 kilometres east in 2 hours and what will happen to the time and 72 in 1 hour what is the displacement you after take the displacement formula that is total time taken divided by the distance travelled ok displacement and distance travelled is different about its terms ok

8 0

3 years ago

A 1.5 kg spherical ball is has a radius of 50 cm is rotating with angular velocity of 12 revolutions per minute. Determine the r

kykrilka [37]

Answer:

K.E = 0.0075 J

Explanation:

Given data:

Mass of the ball = 1.5 kg

radius, r = 50 cm = 0.5 m

Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec

Now,

the kinetic energy is given as:

K.E = $K.E=\frac{1}{2}I\omega^2$

where,

I is the moment of inertia = mr²

on substituting the values, we get

$K.E=\frac{1}{2}\times1.5\times0.5^2\times0.2^2$

K.E = 0.0075 J

3 0

3 years ago