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ipn [44]
3 years ago
9

Two forces of 83 pounds and 56 pounds act simultaneously on an object. The resultant forms an angle of 52° with the 56-pound for

ce. Find the angle formed between the two original forces, to the nearest 10th of a degree.

Physics
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

95.9°

Explanation:

The diagram illustrating the action of the two forces on the object is given in the attached photo.

Using sine rule a/SineA = b/SineB, we can obtain the value of B° as shown in the attached photo as follow:

a/SineA = b/SineB,

83/Sine52 = 56/SineB

Cross multiply to express in linear form

83 x SineB = 56 x Sine52

Divide both side by 83

SineB = (56 x Sine52)/83

SineB = 0.5317

B = Sine^-1(0.5317)

B = 32.1°

Now, we can obtain the angle θ, between the two forces as shown in the attached photo as follow:

52° + B° + θ = 180° ( sum of angles in a triangle)

52° + 32.1° + θ = 180°

Collect like terms

θ = 180° - 52° - 32.1°

θ = 95.9°

Therefore, the angle between the two forces is 95.9°

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Answer:

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Explanation:

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Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

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Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance be
d1i1m1o1n [39]
Use the Inverse square law, Intensity (I) of a light is inversely proportional to the square of the distance(d).

I=1/(d*d)

Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.

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L1/15=(200*200)/(400*400)
L1=15*0.25
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3 years ago
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What causes earthquakes,tsunami, and volcanic eruptions to happen certain places and not others
Dafna11 [192]
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5 0
3 years ago
A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she
tester [92]

Answer:

a)  I = 1,75 10-² kg m²  and b)  I = 1.49 10⁻² kg m²

Explanation:

The expression for the moment of inertia is

    I = ∫ r² dm

The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.

    I = I core + I shell

The moment of inertia of a solid sphere is

    I sphere = 2/5 MR²

The moment of inertia of a thin spherical shell is

    I shell = 2/3 M R²

a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter

     R = d / 2

     R= 0.196 m / 2 = 0.098 m

     I core = 2/5 1.6 0.098²

     I core = 6.147 10-3 kg m²

Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m

    R = 0.206 / 2

    R = 0.103 m

    I shell = 2/3 1.6 0.103²

    I shell = 1,132 10-2 kg m²

The moment of inertia of the ball is the sum of these moments of inertia,

    I = I core + I shell

    I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³

    I = 17.47 10⁻³ kg m²

    I = 1,747 10-² kg m²

b) Now the ball is report with mass 3.2kg and diameter 0.216 m

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    R = 0.108 m

It is a uniform sphere

    I = 2/5 M R²

    I = 2/5 3.2 0.108²

    I = 1.49 10⁻² kg m²

7 0
3 years ago
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