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3 years ago

Find the average net force.

Physics

1 answer:

Lerok [7]3 years ago

6 0

Answer:

<h2>given</h2>

mass= 3.5 kg

distance covered= 0.4m

inital speed = 1.5 m/sec

final speed = 0 m/sec

net force

<h2>solution</h2>

formula for force :

$\fbox{f = m.a}$

In order to find the force we need to find the acceleration.

According to third equation of Kinematics-

$\bold{ {v}^{2} = {u}^{2} + 2as}$

where,

v= final velocity

u=inital Velocity

a=Acceleration

s=Distance covered

put the value in the above equation

0= (1.5)²+ 2a x 0.4

-2.25= 0.8a

a = -2.25/0.8

a= - 2.8125 m/sec²

now put the value 0f accleration in formula of force

F = 3.5 x - 2.8125

F= -9.84375 N

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1) Consider a source particle of charge qS=9 C located at (−9,5) [distances in meters]. Find the electric field vector at the ta

xeze [42]

Answer:

1) E = 2.25 i^+ 0.809j^) 10⁹ N / C , 2) E = 2.39 10⁹ N / C , 3) θ = 19.8º , 4) F = 19.12 10⁹ N , 5) E = (1.32 i^+ 3.56 j^) 109 N/C

Explanation:

1) The equation for the electric field is

E = k q / r²

Where K is the Coulomb constant that is worth 8.99 10⁹ N m² /C², q is the load and r is the distance of the load to the test point

Since the electric field is a vector magnitude, we can find its component

X axis

Ex = k q / x²

where the distance on the axis is

x = √ (X₂-x₁)²

x = √ (-15 + 9)² = 6 m

Eₓ = 8.99 10⁹ 9/6²

Eₓ = 2.25 10⁹ N /C

Y Axis

y = √ (y₂-y₁)² = √ (15-5)² = 10 m

$E_{y}$ = 8.99 10⁹ 9/10²

$E_{y}$ = 0.809 10⁹ N / C

E = Eₓ i^+ $E_{y}$ j^

E = 2.25 i^+ 0.809j^) 10⁹ N / C

2) the magnitude can be found using the Pythagorean triangle

E = √ (Eₓ² + $E_{y}$ ²)

E = √ (2.25² + 0.809²) 10⁹

E = 2.39 10⁹ N / C

3) to find the angle let's use trigonometry

tan θ = $E_{y}$ / Eₓ

θ = tan⁻¹ $E_{y}$ / Eₓ

θ = tan⁻¹ (0.809 / 2.25)

θ = 19.8º

Regarding the positive side of the x axis

4) a charge q2 = 8C is placed, let's calculate the force

F = q E

F = 8 2.39 10⁹

F = 19.12 10⁹ N

5) The total electric field at the origin, let's look for its components

q₁ = 9C

r₁ = -9 i ^ + 5 j ^

q₂ = 8 C

r₂ = -15 i ^ + 15 j ^

X axis

Eₓ = E₁ₓ + E₂ₓ

Eₓ = k q₁ / Dx₁² + k q₂ / Dx₂²

Eₓ = 8.99 10⁹ (9 / (9-0)² + 8 / (15-0)²)

Eₓ = 1.32 109 N / A

Y Axis

$E_{y}$ = $E_{1y}$ + $E_{2y}$

$E_{y}$ = k q₁ / Δy₁² + k q₂ / Δy₂²

$E_{y}$ = 8.99 109 (9/5² + 8/15²)

$E_{y}$ = 3.56 109 N / A

E = (1.32 i^+ 3.56 j^) 109 N/C

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4 years ago

A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.

vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

F = B i L

Rod is not necessarily vertical

$F_x =i L B_d$

$F_y= i L B_w$

the normal reaction N = mg-F y

static friction f = μ_s (mg-F y )

horizontal acceleration is zero

$F_x-f = 0$

$iLBd = \mu_s(mg-F_y )$

B_w = B sinθ

B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

$B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}$

$\theta =tan{-1}{\mu_s}$

$\theta =tan{-1}{0.6}$

$\theta = 31^0$

$B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}$

B = 0.1 T

4 0

4 years ago

2. Which is NOT a unit of speed?

Ratling [72]

The answer is D) second/meter

7 0

2 years ago

If two objects A and B have the same kinetic energy but A has three times the momentum of B, what is the ratio of their inertias

Thepotemich [5.8K]

Answer:

$\frac{inertia_B}{inertia_A}=9$

Explanation:

First of all, let's remind that:

- The kinetic energy of an object is given by $K=\frac{1}{2}mv^2$ , where m is the mass and v is the speed

- The momentum of an object is given by $p=mv$

- The inertia of an object is proportional to its mass, so we can write $I=km$ , where k just indicates a constant of proportionality

In this problem, we have:

- $K_A = K_B$ (the two objects have same kinetic energy)

- $p_A = 3 p_B$ (A has three times the momentum of B)

Re-writing both equation we have:

$\frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2\\m_A v_A = 3 m_B v_B$

If we divide first equation by second one we get

$v_A = 3 v_B$

And if we substitute it into the first equation we get

$m_A (3 v_B)^2 = m_B v_B^2\\9 m_A v_B^2 = m_B v_B^2\\m_B = 9 m_A$

So, B has 9 times more mass than A, and so B has 9 times more inertia than A, and their ratio is:

$\frac{I_B}{I_A}=\frac{km_B}{km_A}=\frac{9m_A}{m_A}=9$

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3 years ago

3. How do you represent the strength of a force in a free-body diagram?

ivann1987 [24]

Generally, the length of the line will indicate how strong the force is. If you have two opposing forces and one is higher than the other, you would draw the line of the higher force visibly longer.

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4 years ago