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NNADVOKAT
13 days ago
14

Greek engineers had the unenviable task of moving large columns from the quarries to the city. One engineer, Chersiphron, tried

several different techniques to do this. One method was to cut pivot holes into the ends of the stone and then use oxen to pull the column. The 4-ft diameter column weighs 13200 lbs, and the team of oxen generates a constant pull force of 1500 lbs on the center of the cylinder G. Knowing that the column starts from rest and rolls without slipping, determine the velocity of its center G after it has moved 5 ft and the minimum static coefficient of friction that will keep it from slipping.

Physics
1 answer:
Yuliya22 [3.3K]13 days ago
7 0
This should cover all your needs:)
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When Royce was 10 years old, he had a mass of
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The gravitational force between Royce and Earth would double by the age of 16. Newton’s law of universal gravitation states that the gravitational force is proportional to the masses involved and inversely proportional to the square of the distance between them. At age 10, Royce weighed 30 kg, and by age 16, he weighed 60 kg. Since his mass doubled from 10 to 16 years, this results in a corresponding doubling of the gravitational force.
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23 days ago
Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle
Ostrovityanka [3204]

Answer:

Explanation:

Let us denote the launch angle as \theta _1=30^{\circ}.

\theta _2=60^{\circ}, \theta _1, and \theta _2 are complementary angles which means their ranges are identical.

H_{max}=\frac{u^2(\sin \theta )^2}{2g}

H_{30}=\frac{u^2(\sin 30)^2}{2g}

H_{30}=\frac{u^2}{8g}

Time of flight is indicated as =\frac{2u\sin \theta }{g}.

t_{30}=\frac{u}{g}

For \theta =60^{\circ}

H_{60}=\frac{u^2(\sin 60)^2}{2g}

H_{60}=\frac{3u^2}{8g}

t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}

H_{60}=3\times H_{30}

t_{60}=\sqrt{3}\times t_{30}

7 0
1 month ago
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Softa [3030]

Answer:

H = 10.05 m

Explanation:

The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s

therefore, the total duration of the upwards motion above the peak of the pole is provided as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

this indicates the speed at the flagpole's top

at this point we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

the height of the flagpole is stated as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
2 months ago
Which statement is always false for athletes participating in team sports?
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Which statement can never be true for athletes in team sports? The statement that is always false among the listed options for team sports athletes is choice C) Conflict resolution indicates a lack of sportsmanship. The other statements are valid in the context of team sports.
3 0
1 month ago
A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4
Yuliya22 [3333]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

R=\sqrt{P^2+Q^2+2PQCos\theta}

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

\theta= 90^o

R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70

(Cos90^o=0),(sin 90^o=1)

\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0
2 months ago
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