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Over [174]
3 years ago
13

The type of brightness in which all

Physics
1 answer:
Gnoma [55]3 years ago
3 0
The answer is C . obvious brightness
You might be interested in
What qualities did Galilea, Sir Isaac Newton, and Albert Einstein have?
MA_775_DIABLO [31]
The qualities that Galileo , Sir Isaac Newton, and Albert Einstein is that they all created models of nature. Galileo discovered Jupiter's four moons and declared that the earth revolves around the sun.Newton developed the three laws of motion, which formed the basic principles of modern physics.Albert Einstein discovered the general theory of relativity, but best known for his mass-energy equivalence formula E=mc².
4 0
3 years ago
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
2 years ago
Which of the following examples could this free body diagram describe? (Check all that apply)
boyakko [2]

Answer:

A car accelerating to the right

Explanation:

The free-body diagram shows all the forces acting on an object. The length of each arrow is proportional to the magnitude of the force represented by that arrow.

In this free-body diagram, we see that there are 4 forces acting on the object, in 4 different directions. We also see that the two vertical forces are equal so they are balanced, while the force to the rigth is larger than the force to the left: this means that there is a net force to the right, so the object is accelerating to the right.

Therefore, the correct answer is:

A car accelerating to the right

5 0
3 years ago
Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65
guajiro [1.7K]

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, \theta =65°

Range of the projectile is, R=520 m.

Acceleration due to gravity, g=9.8\ m/s^2

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

R=\frac{v_{0}^2\sin2\theta}{g}

Plug in all the given values and solve for minimum speed, v_0.

520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

3 0
3 years ago
The diagram below shows some waves traveling along a rope.
Harrizon [31]

-- Well first of all, the waves are <em>transverse waves</em>.  The fibers in the rope, and the hand that generates the waves, are moving up and down, but the wave is moving to the right, towards the wall.  These directions are perpendicular.

-- Later on, after the waves reflect from the wall and travel back toward the hand, there are going to be <em>standing waves</em> on the rope.  But this is probably beyond the scope of the question.

3 0
3 years ago
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