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Romashka [77]
2 years ago
11

The following coplanar forces pull on a ring 200N at 30 degrees and 500N at 80 degrees and 300N at 240 degrees and an unknown fo

rce. Find the magnitude and direction of the unknown force of the ring is to be in equilibrium
Physics
1 answer:
Alex73 [517]2 years ago
3 0

MAGNITUDE AND DIRECTION OF A VECTOR

Given a position vector →v=⟨a,b⟩,the magnitude is located by |v|=√a2+b2. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tanθ=(ba)⇒θ=tan−1(ba)

<h3>What is the formula of magnitude?</h3>

The formula to determine the extent of a vector (in two dimensional space) v = (x, y) is: |v| =√(x2 + y2). This formula is derived from the Pythagorean theorem. the procedure to determine the magnitude of a vector (in three dimensional space) V = (x, y, z) is: |V| = √(x2 + y2 + z2)

To learn more about magnitude and direction, refer

brainly.com/question/27248532

#SPJ9

main answer is given below,

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Answer:

call 7194936846 and ill help u out

Explanation:

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Copper exists in nature as two isotopes. The atomic masses and relative abundance of these isotopes is given in the table. What
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<span>The correct answer should be B) 63.55. That's because the most precise number is 63.546, but you would write 55 because 46 is rounded that way in the equation. The others are a bit higher, while E is a completely different element, Iodine. This isn't the most precise piece of data because in reality there would be a slight differentiation of +- 0,003u</span>
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3 years ago
Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi
Oksanka [162]
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>
6 0
3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him
juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal  direction

m_{b}v_{b}\cos\theta= m_{c}v_{c}

v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}

Put the value into the formula

v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}

v_{c}=0.0345\ m/s

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

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3 years ago
The momentum of a 5-kilogram object moving at 6 meters per second is-
goldfiish [28.3K]
I think the answer is 30 but I’m not sure
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