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Ainat [17]
3 years ago
13

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at

an angle of ? = 57.0 ? above the negative xaxis in the second quadrant. F? 2 has a magnitudeof 5.20 N and is directed at an angle of ? = 53.7 ? below the negative x axis in the third quadrant.
Part A

What is the x component Fx of the resultant force?

Express your answer in newtons.

Part B

What is the y component Fy of the resultant force?

Express your answer in newtons.

Part C

What is the magnitude F of the resultant force?

Express your answer in newtons.

Part D

What is the angle ? that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.
Physics
1 answer:
Stolb23 [73]3 years ago
7 0

Answer:

a. Fx = -8.089 N b. Fy = 3.525 N c. 8.824 N d. 336.45°

Explanation:

Since F₁ = 9.2 N and acts at 57° above the negative axis in the second quadrant, its x-component is -F₁cos57° and its y- component is F₁sin57°

Since F₁ = 5.2 N and acts at 53.7° below the negative axis in the third quadrant, its x-component is -F₂cos53.7° and its y- component is -F₂sin53.7°

Part A

What is the x component Fx of the resultant force?

The x component of the resultant force Fx = -F₁cos57° + -F₂cos53.7° = -9.2cos57° + (-5.2cos53.7°) = (-5.011 - 3.078) N = -8.089 N

Part B

What is the y component Fy of the resultant force?

The y component Fy of the resultant force = F₁sin57° + -(F₂sin53.7°) = 9.2sin57° - 5.2sin53.7° = (7.716 - 4.191) N = 3.525 N

Part C  

What is the magnitude F of the resultant force?

The magnitude F of the resultant force = √(Fx² + Fy²)

F = √(-8.089² N + 3.525² N) = √65.432 + 12.426 = √77.858 = 8.824 N

Part D

What is the angle ? that the resultant force forms with the negative x axis?

The angle the resultant force makes with the negative x axis is given by

θ = tan⁻¹(Fy/Fx) = tan⁻¹(3.525/-8.089) = tan⁻¹-0.4358 = -23.55°.

To measure it from the negative x axis, we add 360. So, our angle = 360 -23.55 = 336.45°

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Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

v =\sqrt{\dfrac{T}{\mu}}

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}

   v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

          λ=2 L

          λ=2 x 0.577 = 1.154 m

we now,

       v = f λ

      f = \dfrac{v}{\lambda}

      f = \dfrac{1074.49}{1.154}

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7 0
3 years ago
The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The r
ira [324]

Given Information:

Number of turns = N = 84

Area of Rectangular coil = 2.61x3.64 cm = 0.0261x0.0364 m

Magnetic field = B = 0.80 T

Current = I = 10.5 mA = 0.0105 A

Angular speed = ω = 3.54x10³ rev/min

Required Information:

(a) Maximum torque = τmax = ?

(b) Peak output power = Ppeak = ?

(c) Work done = W = ?

(d) Average power = Pavg?

Answer:

(a) Maximum torque = 0.00067 N.m

(b) Peak output power = 0.248 W

(c) Work done = 0.00189 J

(d) Average power = 0.1115 W

Explanation:

(a) The toque τ acting on the rotor is given by,

τ = NIABsin(θ)

Where N is the number of turns, I is the current, A is the area of rectangular coil and B is the magnetic field

A = 0.0261x0.0364

A = 0.00095 m²

The maximum toque τ is achieved when θ = 90°

τmax = NIABsin(90°)

τmax = 84*0.0105*0.00095*0.80*1

τmax = 0.00067 N.m

(b) The peak output power of the motor is given by,

Pmax = τmax*ω

ω = 3.54x10³ x 2π/60

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Pmax = 0.00067*370.7

Pmax = 0.248 W

(c) The amount of work done by the magnetic field on the rotor in every full revolution is given by

W = 2∫NIABωsin(ωt) dt

W = -2NIABcos(ωt)

Evaluating limits,

W = -2NIABcos(π) - (-2NIABcos(0))

W = 2NIAB + 2NIAB

W = 4NIAB

W = 4*84*0.0105*0.00067*0.80

W = 0.00189 J

(d) Average power of the motor is given by

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t = 2π/370.7

t = 0.01694 sec

Pavg = W/t

Pavg = 0.00189/0.01694

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Answer:

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This is Fabric B. Wool is basically hard to ignite. It burns slowly. Its ash is dark in color. The wool fabric smells of burning hair when it is burnt.

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This is Fabric C. The silk yarn shrinks from the flame when it is burned. It has a grayish black ash. The silk yarn fabric smells burnt meat of  when it is burnt.

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Anika [276]

Answer:

1885.2 ohms

Explanation:

Step one:

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L=5H

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