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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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A boy throws a ball of mass 0.22 kg straight upward with an initial speed of 29 m/s. When the ball returns to the boy, its speed
maksim [4K]

Answer:

The work is -67.76 J

Explanation:

The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.

This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.

In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.

So, the loss of kinetic energy is \frac{1}{2} *m*(vf^{2} -vi^{2} )

You know:

  • mass=m=0.22 kg
  • Initial velocity of the ball: vi= 29 \frac{m}{s}

Final velocity of the ball: vf= 15 \frac{m}{s}

Replacing:

\frac{1}{2} *0.22 kg*(15^{2} -29^{2} )= -67.76 J

Friction work is always negative because friction is always against displacement.

<u><em>The work is -67.76 J</em></u>

5 0
3 years ago
If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o
andrezito [222]
The answer is

18 / x = 12 / 4 
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a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector
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Answer:

magnitude=34.45 m

direction=55.52\°

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance d between two points:

d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}} (1)

d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}} (2)

d=\sqrt{1186.81 m^{2}} (3)

d=34.45 m (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

tan \theta=\frac{Y2-Y1}{X2-X1}  (5)

tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}  (6)

tan \theta=\frac{24.8}{19.5}  (7)

Finding \theta:

\theta= tan^{-1}(\frac{24.8}{19.5})  (8)

\theta= 55.52\°  (9) This is the direction of the vector

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