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garri49 [273]
3 years ago
5

A ball is thrown at a 60.0° angle above the horizontal across level ground. It is thrown from a height of 2.00 m above the groun

d with a speed of 20.0 m/s and experiences no appreciable air resistance. The time the ball remains in the air before striking the ground is closest to_____________.a. 16.2 s.b. 3.07 s.c. 3.32 s.d. 3.53 s.e. 3.64 s.

Physics
1 answer:
klio [65]3 years ago
7 0

Answer:

Option E is correct.

Time the ball remains in the air before striking the ground is closest to 3.64 s

Explanation:

yբ = yᵢ + uᵧt + gt²/2

yբ = 0

yᵢ = 2 m

uᵧ = u sinθ = 20 sin 60 = 17.32 m/s

g = -9.8 m/s², t = ?

0 = 2 + 17.32t - 4.9t²

4.9t² - 17.32t - 2 = 0

Solving the quadratic equation,

t = 3.647 s or t = -0.1112 s

time is a positive variable, hence, t = 3.647 s. Option E.

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A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.
spayn [35]

Answer:

5.09 m/s

Explanation:

Use the height to find the time it takes to land:

y = y₀ + v₀ᵧ t + ½ gt²

0 = 8.0 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.28 s

Now use the horizontal distance to find the initial velocity.

x = x₀ + v₀ₓ t + ½ at²

6.5 m = 0 m + v₀ (1.28 s) + ½ (0 m/s²) (1.28 s)²

v₀ = 5.09 m/s

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Refraction occurs when __________.
ser-zykov [4K]
Refraction occurs when a wave enters a new medium and changes its speed.

i hope this helps 
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Solve the inequality. x/3 is greater than or equal to - 6. a. x ≥ –9 b. x ≥ 9 c. x ≥ –18 d. x ≤ –18
NISA [10]
To get x on its own, you times the 3 over to the other side so the 3 cancels out on the LHS. 

~ x greater than or equal to -18

(C)
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You blow across the open mouth of a test tube and produce the fundamental standing wave of the air column inside the test tube i
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B: 980 Hz

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Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. the tide measures 5.15 ft at
RSB [31]
<span>f(x) = 5.05*sin(x*pi/12) + 5.15

   First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.

   So our function at this point looks like f(x) = 5.05*sin(x*pi/12) But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
  f(x) = 5.05*sin(x*pi/12) + 5.15

   The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
 f(x) = 5.05*sin(x*pi/12 + C) + 5.15
  where
 C = Phase correction offset.

   But we don't need it for this problem, so the answer is:
 f(x) = 5.05*sin(x*pi/12) + 5.15

   Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>
7 0
3 years ago
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