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Svetach [21]
3 years ago
14

Muscles that are fatigued ____. A. Are injured , B. Should always be rested , C. Can sometimes be exercised further D. Should be

stretched
Physics
2 answers:
Lorico [155]3 years ago
5 0

Answer:

<u>B. Should be rested</u>

Explanation:

That is the only reasonable answer.

Anestetic [448]3 years ago
4 0

Answer:

d

Explanation:

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Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s
alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

  • Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
  • Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
  • vₓ₀ = v * cos θ (1)
  • where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
  • Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

  • At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
  • Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
  • vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

  • At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

  • Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
  • v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
  • Replacing (7) in (6), we get:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

e)

  • When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
  • The horizontal component, since it keeps constant, is just v₀x:
  • v₀ₓ = 13.7 m/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

3 0
3 years ago
How do you do this?Or what are the formulus
LuckyWell [14K]
You got the formulas on the sheet on the top :) So just use those, exchanging v (as in velocity, expressed in m/s) and the d (in meters) and t (in seconds). Hope you will manage it.
4 0
3 years ago
1.A student attaches a string to a puck on a frictionless air table, and pulls with a constant force on the puck. a. Draw the fo
Mrrafil [7]

a)

for the puck :

F = force applied in the direction of pull

N = normal force on the puck in upward direction by the surface of table

W = weight of the puck in down direction due to force of gravity


b)

along the vertical direction , normal force balance the weight of the puck , hence the net force is same as the force of pull F .

so  F = ma                                    where m = mass of puck  , a = acceleration

Fnet = F


c)

since the net force acts in the direction of force of pull F , hence the puck accelerates in the same direction .

6 0
3 years ago
A book is sitting on a desk. What best describes the normal force acting on the book?​
maxonik [38]

Gravitational force is the result of the earth pulling down on the book, so the normal force is best described as the earth or more accurately the table pushing up on the book

8 0
3 years ago
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When is the next total solar eclipse in north america
matrenka [14]

April 8, 2024

The path of totality will cross Mexico, the USA, and Canada.

5 0
3 years ago
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