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KIM [24]
3 years ago
11

Find the resultant force And angle

Physics
1 answer:
Oksanka [162]3 years ago
5 0

The first force has magnitude 60 N and makes an angle of -30° relative to the positive horizontal axis (i.e. to the immediate right), while the second force has magnitude 150 N and direction +30° from the horizontal.

In component form, we have

\vec F_1 = F_{1,x}\,\vec\imath + F_{1,y}\,\vec\jmath \\\\ \vec F_1 = (60\,\mathrm N)\cos(-30^\circ)\,\vec\imath + (60 \,\mathrm N)\sin(-30^\circ)\,\vec\jmath \\\\ \vec F_1 \approx (52.0\,\mathrm N)\,\vec\imath - (30\,\mathrm N)\,\vec\jmath

\vec F_2 = F_{2,x}\,\vec\imath + F_{2,y}\,\vec\jmath \\\\ \vec F_2 = (150\,\mathrm N)\cos(30^\circ)\,\vec\imath + (150\,\mathrm N)\sin(30^\circ)\,\vec\jmath \\\\ \vec F_2 \approx (130\,\mathrm N)\,\vec\imath + (75\,\mathrm N)\,\vec\jmath

The resultant force is the vector

\vec F = \vec F_1 + \vec F_2 \\\\ \vec F \approx (182\,\mathrm N)\,\vec\imath + (45\,\mathrm N)\,\vec\jmath

Its magnitude is

R = \|\vec F\| \\\\ R \approx \sqrt{(182\,\mathrm N)^2 + (45\,\mathrm N)^2} \\\\ R = 187.35\,\mathrm N \approx \boxed{190\,\mathrm N}

Its direction \theta is such that

\tan(\theta) \approx \dfrac{45\,\mathrm N}{182\,\mathrm N}

Because the components of the resultant force are both positive, that means the angle \vec F makes with the horizontal is between 0° and 90°. So

\theta \approx \tan^{-1}\left(\dfrac{45\,\mathrm N}{182\,\mathrm N}\right) \approx \boxed{14^\circ}

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alexgriva [62]

Answer:F=1.7802

Explanation:

Since we've been given the mass to be .18kg,we are asked to find the Force of which the formulae is

F=ma where f-force,m-mass and a-acceleration due to gravity

So we can just substitute

F-?.m=.18 and a9.89

F=.18×9.89

F=1.7802N

6 0
3 years ago
You Discover A Beautiful Island Upon Which You May Build Your Own Society. You Make The Rules. What Is The First Rule You Put In
alekssr [168]

Answer: no plastic

Explanation:

4 0
3 years ago
Read 2 more answers
A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of t
Maslowich

Answer:

a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s

b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s

Explanation:

a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)

It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.

Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).

Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s

b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.

The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,

0.5T = 2.76 s

T = 2 × 2.76 = 5.52 s

Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as

y = A cos wt (provided that there's no phase difference, that is, Φ = 0)

A = amplitude = 0.18 m

w = (2π/5.52) = 1.138 rad/s

When y = 0.09 m, the time = t₁₂ = ?

0.09 = 0.18 Cos 1.138t₁ (angles in radians)

Cos 1.138t₁ = 0.5

1.138t₁ = arccos (0.5) = (π/3)

t₁ = π/(3×1.138) = 0.92 s

When y = -0.09 m, the time = t₂ = ?

-0.09 = 0.18 Cos 1.138t₂ (angles in radians)

Cos 1.138t₂ = -0.5

1.138t₂ = arccos (-0.5) = (2π/3)

t₂ = 2π/(3×1.138) = 1.84 s

Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s

Hope this Helps!!!

3 0
3 years ago
Please someone do this <br>please ​
monitta

The various contributions involved till the chapati is made is given below.

<h3>What is food?</h3>

The substance that we intake for the body to charge up by giving nutrients is called the food.

Wheat is a staple food. We make chapati from flour obtained from the wheat grains.

The various contributions involved till the chapati is made is given below.

                 Take required amount of atta in a container

                                                     ↓

                    Add water accordingly to form a dough

                                                     ↓

                 Apply oil to make dough smooth for long time

                                                    ↓

        Take small dough, make it a ball shaped and apply dry flour

                                                    ↓

               Roll it using rolling pin on the chapati maker plate

                                                    ↓

      After making it circular or any shape you want, place it on hot tawa

                                                    ↓

                               Bake it on both the sides

                                                   ↓

                                      Chapati is ready

Thus, the flow chart is made.

Learn more about food.

brainly.com/question/16327379

#SPJ1

6 0
2 years ago
Suppose a uniform electric field of 4 N/C is in the positive x direction. When a charge is placed at and fixed to the origin, th
Yuliya22 [10]

Answer:

E_total = 3 N / A

Explanation:

The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.

               E_total = E + E₁

the expression for the field of a point charge is

                E₁ = k q₁ / r²

for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative

                 E_total = E -k q₁ / r₂

we substitute

                   0 = E - k q₁ / r²

                   q₁ = \frac{E r^2}{k}

let's calculate

                   q₁ = \frac{4 \ 2^2}{9 \ 10^{-9}}

                   q₁ = 1.78 10⁻⁹ C

now we can calculate the field for position x = 4 m

                   E_total = 4 - 9 10⁹  1.78 10⁻⁹ / 4²2

                   E_total = 3 N / A

8 0
3 years ago
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