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Agata [3.3K]
2 years ago
14

In gym class you run 22m horizontal then climb a rope vertically for 4.8m. What is the direction angle of your total displacemen

t, as measured from the horizontal?

Physics
1 answer:
Tanya [424]2 years ago
8 0

The answer is: 12.30 degrees.

The explanation for this exercise is shown below:

1. You can make a right triangle as you can see in the picture attached, where \alpha is the angle asked, the opposite side is 4.8 meters and the adjacent side is 22 meters.

2. Therefore, you can use tan^{-1} to calculate the angle, as following:

tan^{-1}(\alpha)=\frac{opposite}{adjacent}\\tan^{-1}(\alpha)=\frac{4.8}{22}\\\alpha=12.30degrees

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A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0
2 years ago
Will give brainliest! how does an engineer use physical science?
pentagon [3]

Answer: gravity, circuits

Explanation:

3 0
2 years ago
Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then t
pashok25 [27]

Answer:

Natae Si Jordan Kaya Sya Napaihe

Explanation:

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4 0
2 years ago
A(n) 83 kg fisherman jumps from a dock into a 139 kg rowboat at rest on the West side of the dock. If the velocity of the fisher
Anna71 [15]

here we can say that there is no external force on fisherman and dock

so here we will use momentum conservation theory

As per momentum conservation

initial momentum of fisherman + boat = final momentum of fisherman + boat

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now we will have

83(3.1) + 139(0) = 83 v + 139 v

257.3 = 222v

v = 1.16 m/s

so the speed of boat and fisherman will be 1.16 m/s

3 0
3 years ago
Question 4 (2 points)
Igoryamba

The answer is a) Teres Major Muscle

7 0
1 year ago
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