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lapo4ka [179]
3 years ago
10

Consider two force vectors in the xy-horizontal plane. Suppose a force of 12.7 N pointing along the +x-axis is added to a second

force of 18.1 N directed at 30 degrees to the +x-axis , also in the horizontal plane. Find the resultant vector for this sum. magnitude direction degrees above the +x-axis in the horizontal plane
Physics
1 answer:
emmasim [6.3K]3 years ago
4 0

Answer:

F_1+F_2= (28.26, 9.05) N

\alpha = 17.7\º

F = 29.67 N

Explanation:

Hi!

In a (x, y) coordinate representation, the two forces are:

F_1=(12.7N, 0)\\F_2=(18.1N\cos(30\º), 18.1N \sin(30\º) )\\\cos(\º30)=0.86\\\sin(\º30)= 0.5

The sum of the two forces is:

F_1 + F_2 = ( 12.7 + 0.86*18.1, 18.1*0.5) N

F_1+F_2= (28.26, 9.05) N

The angle to x-axis is calculated using arctan:

\alpha = \arctan(\frac{F_y}{F_x}) = \arctan(\frac{9.05}{28.26} = 17.7\º

The magnitude is:

F = \sqrt {F_x^2 + F_y^2}= \sqrt{798.6 + 81.9} = 29.67 N

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Answer:

750 kg.m/s

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Initial velocity of car v₁=5 m/s

Initial momentum= 100×5= 500 kg.m/s

New mass after picking two extra people m₂= 150 kg

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How do I find the x and y components and the resultant force?​
Dafna11 [192]

Explanation:

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2) the diagonal force times the cosine of the angle it makes gives the horizontal component or x component. do this to get the x component of all the diagonal forces.

add all the x components as well as the horizontal forces together to get the final x component

3)using the triangle of vectors, the resultant force is calculated

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The rate of rotation of the disk is gradually increased. The coefficient of static friction between the coin and the disk is 0.5
MrRissso [65]

Question is not complete and the missing part is;

A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.

Answer:

0.828 m/s

Explanation:

Resolving vertically, we have;

Fn and Fg act vertically. Thus,

Fn - Fg = 0 - - - - eq(1)

Resolving horizontally, we have;

Ff = ma - - - - eq(2)

Now, Fn and Fg are both mg and both will cancel out in eq 1.

Leaving us with eq 2.

So, Ff = ma

Now, Frictional force: Ff = μmg where μ is coefficient of friction.

Also, a = v²/r

Where v is linear speed or velocity

Thus,

μmg = mv²/r

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Making v the subject;

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Plugging in the relevant values,

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v = 0.828 m/s

3 0
3 years ago
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