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Semmy
25 days ago
14

The box of negligible size is sliding down along a curved path defined by the parabola y=0.4x2. When it is a A(xA = 2 m, yA = 1.

6 m), the speed is v = 8 m/s and the increase in speed is dv/dt = 4 m/s2
Physics
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8) A soccer goal is 2.44 m high. A player kicks the ball at a distance 10.0 m from the goal at an angle of 25.0°. The ball hits
Yuliya22 [3333]
The soccer ball's initial speed stands at 16.38 m/s. Given that the vertical distance is y = 2.44 m, the horizontal span x = 10.0 m, and the angle of launch θ = 25.0°. The initial velocity comprises two components, Vₓ and V. The calculations are as follows: Vₓ = V cosθ and V = V sinθ. The formula for horizontal distance becomes x = Vₓt. Since g is deemed 0, we can state that: x = Vₓt or 10 = V cos 25 * t. Solving for V gives us 10 = 0.906V * t, thus V * t = 10 / 0.906 = 11.038 m. Regarding the vertical distance (with g being negative due to the upward movement opposing gravity), we use y = V sinθ * t - 1/2 * g * t². Following through with the calculations leads us to determine that the soccer ball's initial speed is indeed 16.38 m/s.
6 0
2 months ago
A 1200-kilogram space vehicle travels at 4.8 meters per second along the level surface of Mars. If the magnitude of the gravitat
Ostrovityanka [3204]
The vehicle experiences a normal force of 4440 N. The normal force acts perpendicular to the ground surface. Key details include the vehicle's mass of 1200 kg and the gravitational force of 3.7 N/kg. We calculate the normal force in Newtons by multiplying these two figures: force = field strength * mass = 3.7 N/kg * 1200 kg = 4440 N.
8 0
2 months ago
Read 2 more answers
Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,
Sav [3153]

Answer:

All three pendulums will have the same angular frequencies.

Explanation:

For a simple pendulum, the time period using the approximation sin(\theta )\approx \thetais expressed as:

T=2\pi\sqrt{\frac{l}{g}}

The angular frequency \omega is defined as

\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}

Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.

6 0
2 months ago
According to a newspaper account, a paratrooper survived a training jump from 1200 ft when his parachute failed to open but prov
Softa [3030]
Let's consider a few possibilities. 1. The lowest velocity of the paratrooper would be just before hitting the ground. 2. Given that the jump originated from a relatively short height, the paratrooper utilized a static line, allowing the parachute to deploy almost instantly after leaping. Hence, we will convert 100 mi/h to ft/s: 100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec. Based on the first assumption, the maximum distance fallen by the paratrooper would equate to 8 seconds at 146.67 ft/s, translating to 8 s * 146.67 ft/s = 1173.36 ft. This calculated distance is nearly on par with the jump height, validating both assumptions 1 and 2. Thus, this scenario seems plausible. Moreover, considering the terminal velocity for a parachutist in a freefall position with limbs spread out typically reaches 120 mi/h, which is slightly above the 100 mi/h mentioned in the article. This as well aligns with the notion of the parachute acting like a flag, adding some air resistance.
7 0
2 months ago
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