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vlabodo [156]
2 years ago
12

The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply

upward at the end. The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. In the women's competition, the end of a typical launch ramp is directed 61° above the horizontal. With this launch angle, a skier attains a height of 16 m above the end of the ramp. What is the skier's launch speed?

Physics
1 answer:
kykrilka [37]2 years ago
4 0

Answer: Got It!

<em>Explanation:  </em>let s = speed at launch

v = 0 at top = s sin 63 - g t

so at top

t = s sin 63/g = .0909 s

h = 13.6 = s sin 63 t - 4.9 t^2

13.6 = .081s^2 - .0405 s^2

s^2 = 336

s = 18.3 m/s

0  0

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Answer:

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2 years ago
If the magnitude of a charge is half as much as another charge, but the force experienced is the same, then the electric field s
Kazeer [188]

Answer:

the electric field strength of this charge is two times the strength of the other charge

Explanation:

Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;

E1/E2 = Q2/Q1

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8 0
2 years ago
5)
miskamm [114]
B . I hope this is right
4 0
3 years ago
An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec
aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

x = V*t = -8\frac{m}{s} *3s = -24m

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

v^2 = v_0^2 + 2a*d

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s

Then, the time of the second phase will be:

t_2 = 9s - t_1 = 9s - 1.143s = 7.857s

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

x = \frac{1}{2}a*t^2 + v_0*t + x_0

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m

So, the total distance covered by this object in meters will be the sum of all the distances we found:

x_total = 24m + 4.57m + 216.07m = 244.64m

8 0
2 years ago
2. What do pitch and loudness have in common?
DanielleElmas [232]

Answer:

Both are subject to a persons interpretation

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2 years ago
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