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Nina [5.8K]
3 years ago
14

A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined r

amp which ends 10 feet above a level landing zone. Assume the cyclist maintains a constant speed up the ramp and the ramp is inclined Ao (degrees) above horizontal. With the pictured imposed coordinate system, the parametric equations of the cyclist will be: x(t) = 100t cos(A) y(t) = –16t2 + 100t sin(A) + 10.
Calculate the horizontal velocity of the cyclist at time t; this is the function x'(t) = _______
What is the horizontal velocity if A = 20 degrees?
What is the horizontal velocity if A = 45 degrees? (Four decimal places.)
Calculate the vertical velocity of the cyclist at time t; this is the function y(t) =________.
What is the vertical velocity if A = 20 degrees? (Four decimal places.)
What is the vertical velocity if A = 45 degrees? (Four decimal places.)
The vertical velocity of the cyclist is zero at time ________ seconds.
If the cyclist wants to have a maximum height of 35 feet above the landing zone, then the required launch angle is A = _______ degrees. (Accurate to four decimal places.)
Physics
1 answer:
pantera1 [17]3 years ago
4 0

Answer:

Explanation:

The parametric equations of the cyclist are:

x(t)=100tcos(A)\\\\y(t)=-16t^2+100tsin(A)+10   (1)

A) The horizontal velocity is the derivative of x(t), in time:

x'(t)=100cos(A)

B) For A=20° the horizontal velocity is:

v_x=x'(t)=100cos(20\°)=93.9692ft/s

For A=45°:

v_x=100cos(45\°)=70.7106ft/s

C) To find the time in which the vertical velocity is zero you first obtain the derivative of, in time:

v_y=y'(t)=-32t+100sin(A)+10

Next, you equal the vertical velocity to zero and solve for time t:

-32t+100sin(A)+10=0\\\\t=\frac{100sin(A)+10}{32}

D) The maximum height is reached when the derivative of y (height) is zero. You use the previous value of t in the equation (1), equals y to 35. Next, you  solve for t:

y=35\\\\-16(\frac{100sin(A)+10}{32})^2+100(\frac{10sin(A)+10}{32})sinA+10=35\\\\-\frac{16}{1024}(10000sin^2A+2000sinA+100)+31.25sin^2A+31.25sinA+10=35\\\\-156.25sin^2A-31.25sinA-1.5625+31.25sin^2A+31.25sinA+10=35\\\\-125sin^2A-26.5625=0

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PART ONE
Helga [31]

Answer:

1.129×10⁻⁵ N

1.295 m

Explanation:

Take right to be positive.  Sum of forces on the 31.8 kg mass:

∑F = GM₁m / r₁² − GM₂m / r₂²

∑F = G (M₁ − M₂) m / r²

∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²

∑F = 1.129×10⁻⁵ N

Repeating the same steps, but this time ∑F = 0 and we're solving for r.

∑F = GM₁m / r₁² − GM₂m / r₂²

0 = GM₁m / r₁² − GM₂m / r₂²

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

516 / r² = 207 / (0.482 − r)²

516 (0.482 − r)² = 207 r²

516 (0.232 − 0.964 r + r²) = 207 r²

119.9 − 497.4 r + 516 r² = 207 r²

119.9 − 497.4 r + 309 r² = 0

r = 0.295 or 1.315

r can't be greater than 0.482, so r = 0.295 m.

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3 years ago
A bicyclist covers the first leg of a journey that is d1meters in t1seconds at a speed of v1m/s and the second leg of d2meters i
Murljashka [212]

According to motion in straight line  t1≠t2


A biker travels d1 meters in t1 seconds at v1 m/s for the first leg and d2 meters in t2 seconds at v2 m/s for the second leg. It's possible that t1t2 if his average speed is equal to the average of v1 and v2.
An object is said to be in motion if its position in relation to its surroundings changes over time. It is a shift in an object's position over time. The only type of motion that exists is motion in a straight line.


A reference system is constantly used to describe a particle's motion. An arbitrary origin point is used to create a reference system, and a coordinate system is imagined to be connected to it. The reference system for a specific problem is the coordinate system that has been selected for it. For the majority of the problems, we typically select an earth-based coordinate system as the reference system.


To learn more about Motion in straight lines please visit -brainly.com/question/17675825
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4 0
2 years ago
Einstein’s general theory of relativity made or allowed us to make predictions about the outcome of several experiments that had
PtichkaEL [24]

Answer:

1. bending of light in gravitational fields.

2. effect of gravitational redshift.

3. perihelion precission of mecury.

Explanation:

1 bending of light in gravitational fields, we can think of it like this:

by noting the change in position s of stars as they pass near the sun on the celetial sphere, so since the sun creates a gravitational field even the star thats not in our line of side(behind the sun) can be seen because its light is bent.

2. effects of gravitational redshift:

this says that if you are in the gravitational field, your clock moves slower when it is seen by a distant observer.

3. perihelion precission of mecury:

according to Newtonian physics a two body system consisting of a lone orbiting the spherical mass would trace out an ellipse with the center of mass of the system as the focus but mercury deviates from that precission. then according to Einstein, the change in orientation of the orbital ellipsewithin its orbital plane is the effect of gravitation being mediated by the curvature of space-time.

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I think the answer is A.........

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How much energy from the sun actually reaches the corn answer?
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The energy from the sun that reaches the corn is about two billionths.
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