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Yanka [14]
3 years ago
11

A truck travels up a hill with a 5.7° incline. The truck has a constant speed of 22 m/s. What is the horizontal component of the

truck’s velocity? Answer in units of m/s.

Physics
1 answer:
tester [92]3 years ago
8 0

Answer:

The horizontal component of the velocity is 21.9 m/s.

Explanation:

Please see the attached figure for a better understanding of the problem.

Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.

To find vx, let´s use the following trigonometric rule of right triangles:

cos α = adjacent / hypotenuse

cos 5.7° = vx / 22 m/s

22 m/s · cos 5.7° = vx

vx = 21.9 m/s

The horizontal component of the velocity is 21.9 m/s.

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A pulley system lifts a 1345 n weight a distance of 0.975m. Paul pulls the rope a distance of 3.90m, exerting a force of 375N. A
Rashid [163]

A. IMA: 4

The Ideal Mechanical Advantage (IMA) is given by:

IMA = \frac{d_i}{d_o}

where

d_i is the input distance

d_o is the output distance

For the pulley system in this problem, d_i = 3.90 m and d_o = 0.975 m, so the IMA is

IMA=\frac{3.90 m}{0.975 m}=4


B. MA: 3.59

The actual mechanical advantage (AMA), or simply the Mechanical Advantage (MA), is given by

MA=\frac{F_o}{F_i}

where F_o is the output force and F_i is the input force. For the pulley system in this problem, F_i = 375 N and F_o = 1345 N, so the MA is

MA=\frac{1345 N}{375 N}=3.59


C. Efficiency: 89.8 %

The efficiency of a machine is equal to the ratio between the MA and the AMA:

\eta = \frac{MA}{AMA} \cdot 100

Therefore, in this case,

\eta=\frac{3.59}{4}\cdot 100=0.898=89.8 \%

3 0
3 years ago
Matter is __________.
dlinn [17]
<span>Matter is c) something that has mass and occupies space. Everything in our known universe takes up space, and everything is made up of matter. Matter as a concept is not something that can have any unit of measurement put upon it. </span>
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By the time she managed/had managed to open the door, the postman already went/had already gone.
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Answer:

<h3>Answer Sentence: </h3><h2><em><u>By the time she had managed to open the door, the postman had already gone</u></em>.</h2>
8 0
3 years ago
Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

Learn more here:

brainly.com/question/20357188

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2 years ago
An object travels 20 m in 10s What is its speed?
ella [17]

speed=distance/time

distance=20m

time=10s

speed=?

speed=20×10

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