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VLD [36.1K]
3 years ago
14

A projectile is fired with an initial speed of 51.2 m/s at an angle of 44.5 degrees above the horizontal on a long flat firing r

ange. Determine the following: a. The maximum height reached by the projectile. b. The total time in the air. c. The total horizontal distance covered (that is, the range)
Physics
1 answer:
irinina [24]3 years ago
3 0
Use the projectile motion equations

H = v^2 x sin^2(θ) ÷ 2g

t = 2 x v x sinθ ÷ g

R = v^2 x sin2θ ÷ g
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I think it's C, three hues that are adjacent on the color wheel
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Accelerating car is a _______
Pavel [41]

Acceleration is the rate of change of velocity as a function of time. For example a car traveling at 50 km/hr starts to accelerate, 10 seconds after, its speed changes to 100 km/hr then the acceleration of the car during the time can be calculated as below: initial speed = 50 km/hr.

5 0
2 years ago
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Before CDs and cassette tapes, there were vinyl records. The most comon spun with an angular velocity of 33.3revolutionsperminut
sashaice [31]

Answer:

539.5°

Explanation:

33.3 revolutions per minute

1 revolution = 360°

1 minute = 60 seconds

hence

33.3 revs ----> 1 minute = 60 seconds

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X = (33.3 x 2.7)÷60 = 1.4985 revolutions in 2.70 seconds

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7 0
3 years ago
this stationary wave is what we call the first harmonic of the first normal mode of the system. in units of l, the length of the
pentagon [3]

First harmonic of a closed pipe is determined as velocity, v, to four times length (4L), F₀ v/4L.

<h3>First harmonic of a closed pipe</h3>

The first harmonic of a closed pipe is the fundamental frequency of the closed of the closed pipe.

L = λ/4

where;

  • L is the length of the pipe
  • λ is the wavelength of sound

λ = 4L

But, v = F₀λ

v = F₀(4L)

F₀ = v/4L

where;

  • F₀ is the first harmonic
  • v is speed of sound

Thus, first harmonic of a closed pipe is determined as velocity, v, to four times length (4L), F₀ v/4L.

Learn more about fundamental frequency here:  brainly.com/question/1967686

#SPJ11

<h3 />
6 0
2 years ago
g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

4 0
3 years ago
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