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Marat540 [252]
2 years ago
10

A 90.0-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from

its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position
Physics
1 answer:
asambeis [7]2 years ago
3 0

Answer:

615 N

Explanation:

If θ is the angle from vertical and T is the rope tension

θ = arcsin (2.0 / 3.5) = 34.85°

Summing vertical forces to zero

Tcosθ - mg = 0

T = 90.0(9.81) / cos34.85 = 1,075.85 N

Summing horizontal forces to zero

F - Tsinθ = 0

F = 1075.85sin34.85 = 1075.85(2.0/3.5) = 614.772... ≈ 615 N

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A solid disk has a mass of 162 kg and a radius of 1.30m. This
RideAnS [48]

Answer:

754.3 m

Explanation:

The moment of inertia of the solid disk:

I = mR^2/2

Where m is the disk mass and R is the radius of the disk.

I = 162*1.3^2/2 = 136.89 kgm^2

The angular kinetic energy of the disk is then:

E_k = I\omega^2/2 = 136.89 * 18^2/2 = 22176.18 J

By law of energy conservation, this energy is converted to potential energy to pick up the 3kg block

let g = 9.8 m/s2

E_p = m_bgh = 22176.18 J

where m_b = 3 kg is the mass of block

3*9.8h = 22176.18

h = \frac{22176.18}{3*9.8} = 754.3 m

8 0
3 years ago
With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how ma
Alenkinab [10]

Answer:

The wavelength stays the same.

Explanation:

When the amplitude is increased, the wavelength stays the same.

Here the wavelength doesn't depend upon the amplitude.

4 0
3 years ago
Which of the following describes the efficiency of real machines?
Leokris [45]
Choice-B is the true one.
3 0
3 years ago
Read 2 more answers
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w
Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

\theta= (0.20 rad/s)(75 s)=15 rad

In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water
sashaice [31]

Answer: the constant angular velocity of the arms is 86.1883 rad/sec

Explanation:

First we calculate the linear velocity of the single sprinkler;

Area of the nozzle = π/4 × d²

given that d = 8mm = 8 × 10⁻³

Area of the nozzle = π/4 × (8 × 10⁻³)²

A = 5.024 × 10⁻⁵ m²

Now total discharge is dived into 4 jets so discharge for single jet will be;

Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec

So using continuity equation ;

Q_single = A × V_single

V_single = Q_single/A

we substitute

V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)

V_single = 29.8566 m/s

Now resolving the forces as shown in the second image,

Vt = Vcos30°

Vt = 29.8566 × cos30°

Vt = 25.8565 m/s

Finally we calculate the angular velocity;

Vt = rω

ω_single = Vt / r

from the given diagram, radius is 300mm = 0.3m

so we substitute

ω_single = 25.8565 / 0.3

ω_single = 86.1883 rad/sec

Therefore the constant angular velocity of the arms is 86.1883 rad/sec

7 0
3 years ago
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