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mylen [45]
3 years ago
7

A machine in an ice factory is capable of exerting 3.31 × 102 N of force to pull large blocks of ice up a slope. The blocks each

weigh 1.32 × 104 N. Assuming there is no friction, what is the maximum angle that the slope can make with the horizontal if the machine is to be able to complete the task? Answer in units of ◦ .

Physics
1 answer:
tankabanditka [31]3 years ago
7 0

Answer:\theta =1.43 ^{\circ}

Explanation:

Given

Force exerted by machine F=3.31 \times 10^2 N

Weight of Block W=mg=1.32\times 10^4 N

Considering inclination to be \theta

After resolving Forces We get

mg\sin \theta =F

\sin \theta =\frac{F}{mg}

\sin \theta =\frac{3.31 \times 10^2}{1.32\times 10^4}

\sin \theta =2.507\times 10^{-2}

\sin \theta =0.02507

\theta =\sin ^{-1}(0.02507)

\theta =1.43 ^{\circ}

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3 years ago
I need help with this physics question ASAP. Thanks!
Ugo [173]

From the calculation, the horizontal distance is 310 m.

<h3>What is the range of a projectile?</h3>

A projectile has to do with an object that is thrown at an angle. Now we know that the range of the projectile is given by the formula; R = u^2sin2θ/g

We we know that;

g = acceleration due to gravity

θ = angle

u = velocity

R = (60)^2 sin2(30)/10

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2 years ago
Okay I better get a good answer for this guy I'm puttin' all my money on it.
Black_prince [1.1K]

Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.

Explanation:

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A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft&gt;s. if the attached cord is pulled
Leya [2.2K]
Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
      = (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer:  44.25 ft-lb


6 0
3 years ago
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