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Bas_tet [7]
4 years ago
11

Joe is attempting to ramp his bicycle over a row of burning tires. If the ramp has a launch angle of 20 degrees, and Joe can ped

al a bicycle at a maximum of 20 km/h, how many burning tires can he jump safely, assuming each tire has a radius of 0.3 meters?
Physics
1 answer:
Luden [163]4 years ago
4 0
Answer is 6 tires.

This is a projectile question.

First make sure units are consistent - express speed in m/s.

20 km/h = 20000m / 3600 s = 5.56 m/s

Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.

Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:

Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s

An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:

Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s

Total trip time:
0.19 x 2 = 0.38s

Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:

v = d / t
5.22 = d / 0.38
d = 1.98m

Now divide this total distance by the length of an individual tire to find the number of tires he will clear:

1.98 / 0.3 = 6.6 tires

Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).

Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!
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In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th
Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0
3 years ago
The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla
Vladimir [108]

Answer

given,

Radius of sphere = 6.38 × 10⁶ m

time  = 1 day = 86400 s

\omega = \dfrac{2\pi}{T}

\omega = \dfrac{2\pi}{ 86400 }

\omega = 7.272 \times 10^{-5}\ rad/s

a) at equator

v = R_E \omega

v = 6.38 \times 10^6\times 7.272 \times 10^{-5}

v = 464 m/s

acceleration of the person

a = \omega^2R_E

a = (7.272 \times 10^{-5})^2(6.38 \times 10^6)

a = 0.03374 m/s^2

b) at a latitude of 61.0 ° north of the equator.

R = R_E cos \theta

R = 6.38 \times 10^6\times cos 61^0

R = 3.093 \times 10^6 m

v = R \omega

v = 3.093 \times 10^6 \times 7.272 \times 10^{-5}

v = 225 m/s

acceleration of the person

a = \omega^2R_E

a = (7.272 \times 10^{-5})^2(3.093 \times 10^6 )

a = 0.01635 m/s^2

4 0
3 years ago
7. A 600g brick weighs 6N in the air. If the brick displaces 2N of water when it is
Oxana [17]
The answer to this would be 12N
6 0
3 years ago
* CRIMINOLOGY*
Setler [38]

I think the answer is B.

Hope this helps.

5 0
4 years ago
Read 2 more answers
If the speed of light is 3.0 × 108 m/s2, then how much energy would be released if a 23.7 gram object is converted to pure energ
elena55 [62]

7.11x 10⁶J

Explanation:

Given parameters:

Speed of light  = 3 x 10⁸m/s

Mass of object = 23.7g = 0.0237kg

Unknown:

Energy = ?

Solution:

From Einstein's equation, we see that mass and energy are equivalent using the expression below:

                                  E = mc²

Substituting the parameters:

                 E = 3 x 10⁸ x 0.0237 = 7.11x 10⁶J

Learn more:

Energy brainly.com/question/5381158

#learnwithBrainly

3 0
4 years ago
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