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3 years ago

The following forces act on an object: 10 N north, 7 N south, and 4 N east. What is the magnitude of the net force?

Physics

1 answer:

spin [16.1K]3 years ago

6 0

Given data;

Fn = 10 N

Fs = 7 N

Fe = 4 N

force in X direction (Fx) = 4 N

force in Y direction (Fy) = 10-7 = 3 N

Net force (Fnet) = Sq.root[(Fx)² + (Fy)²]

= Sq root [ 4² + 3² ]

= 25 N

<em> Net force acting = 25 N</em>

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An astronaut is a short distance away from her space station without a tether rope. She has a large wrench. What should she do w

Hitman42 [59]

Answer: b. Throw it directly away from the space station.

Explanation:

According to <u>Newton's third law of motion</u>, <em>when two bodies interact between them, appear equal forces and opposite senses in each of them.</em>

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Each time a body or object exerts a force on a second body or object, it (the second body) will exert a force of equal magnitude but in the opposite direction on the first.

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3 years ago

Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge

Wittaler [7]

Answer:

F = 0.111015 N

Explanation:

For this exercise the force is given by Coulomb's law

F = k q₁q₂ / r₂₁²

we calculate the electric force of the other two particles on the charge q1

Charges q₁ and q₂

the distance between them is

r₁₂ = y₁ -y₂

r₁₂ = 0.30 + 0.30

r₁₂ = 0.60 m

let's calculate

F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

F₁₂ = 1 10⁻¹ N

directed towards the positive side of the y-axis

Charges 1 and 3

Let's find the distance using the Pythagorean Theorem

r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

r₁₃ = 0.50 m

F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

F₁₃ = 1.697 10⁻² N

The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ 0.3 / 0.4

tea = 36.87º

The angle from the positive side of the x-axis is

θ ’= 180 - θ

θ ’= 180 - 36.87

θ ’= 143.13º

sin143.13 = F_13y / F₁₃

F_13y = F₁₃ sin 143.13

F{13y} = 1.697 10⁻² sin 143.13

F_13y = 1.0183 10⁻² N

cos 143.13 = F_13x / F₁₃

F₁₃ₓ = F₁₃ cos 143.13

F₁₃ₓ = 1.697 10⁻² cos 143.13

F₁₃ₓ = -1.357 10-2 N

Now we can find the components of the resultant force

Fx = F13x + F12x

Fx = -1,357 10-2 +0

Fx = -1.357 10-2 N

Fy = F13y + F12y

Fy = 1.0183 10-2 + 1 10-1

Fy = 0.110183 N

We use the Pythagorean theorem to find the modulus

F = Ra (Fx2 + Fy2)

F = RA [(1.357 10-2) 2 + 0.110183 2]

F = 0.111015 N

Let's use trigonometry for the angles

tan tea = Fy / Fx

tea = tan-1 (0.110183 / -0.01357)

tea = 1,448 rad

to find the angle about the positive side of the + x axis

tea '= pi - 1,448

Tea = 1.6936 rad

6 0

3 years ago

A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak

GarryVolchara [31]

Explanation:

It is given that,

Mass of the box, m = 100 kg

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.

From the attached figure, the x and y component of forces is given by :

$T_{1x} =-T_1 cos (20)$

$T_{2x} = T_2 cos (40)$

$mg_x = 0$

$T_{1y} = T_1 sin (20)$

$T_{2y} = T_2 sin (40)$

$mg_y= -mg$

Let $R_x$ and $R_y$ is the resultant in x and y direction.

$R_x=-T_1 cos (20)+T_2 cos (40)+0$

$R_y=T_1 sin(20)+T_2 sin(40)-mg$

As the system is balanced the net force acting on it is 0. So,

$-T_1 cos (20)+T_2 cos (40)+0=0$ .............(1)

$T_1 sin(20)+T_2 sin(40)-100\times 9.8=0$ ..................(2)

On solving equation (1) and (2) we get:

$T_1=866.86\ N$ (tension on the left rope)

$T_2=1063.36\ N$ (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.

7 0

3 years ago

Four forces are exerted on a disk of radius R that is free to spin about its center, as shown above. The magnitudes are proporti

Dmitry_Shevchenko [17]

The given magnitude of forces of F₁ = F₄, F₂ = F₃, F₁ = 2·F₂, give the

forces that exert zero net torque on the disk as the options;

(B) F₂

(D) F₄

<h3>How can the net torque on the disk be calculated?</h3>

The given parameters are;

F₁ = F₄

F₂ = F₃

F₁ = 2·F₂

Therefore;

F₄ = 2·F₂

In vector form, we have;

$\vec{F_4} = \mathbf{\frac{\sqrt{3} }{2} \cdot F_4 \cdot \hat i - 0.5 \cdot F_4 \hat j}$

$\vec{F_2} = \mathbf{ -F_2 \, \hat j}$

Clockwise moment due to F₄, M₁ = $-0.5 \times F_4 \, \hat j \times \dfrac{R}{2}$

Therefore;

$M_1 =- 0.5 \times 2 \times F_2 \, \hat j \times \dfrac{R}{2} = \mathbf{ -F_2 \, \hat j \times \dfrac{R}{2}}$

Counterclockwise moment due to F₂ = $-F_2 \, \hat j \times \dfrac{R}{2}$

Given that the clockwise moment due to F₄ = The counterclockwise moment due to F₂, we have;

Two forces that combine to exert zero net torque on the disk are;

F₂, and F₄

Which are the options; (B) F₂, and (D) F₄

Learn more about the resolution of vectors here:

brainly.com/question/1858958

4 0

2 years ago