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MariettaO [177]
3 years ago
15

The current world-record motorcycle jump is 77.0 m, set by Jason Renie. Assume that he left the take-off ramp at 11.0° to the ho

rizontal and that the take-off and landing heights are the same. Neglecting air drag, determine his take-off speed.
Physics
1 answer:
Dafna1 [17]3 years ago
7 0

Answer:

44.9 m/s

Explanation:

Initial velocity = v

Angle at which the projectile is shot at = θ = 11°

g = Acceleration due to gravity = 9.81 m/s²

Range of projectile

R=\frac {v^{2}\sin 2\theta}{g}\\\Rightarrow v=\sqrt{\frac{Rg}{sin 2\theta}}\\\Rightarrow v=\sqrt{\frac{77\times 9.81}{sin (2\times 11)}}=44.9\\\Rightarrow v=44.9\ m/s

The take off speed of Jason Renie's motorcycle was 44.9 m/s

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The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i
Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius=r=\frac{d}{2}=\frac{2.3}{2}=1.15 m

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency=\frac{1200}{60}=20 Hz

1 minute=60 seconds

Angular velocity=\omega=2\pi f

Angular velocity=2\times \frac{22}{7}\times 20=125.7 rad/s

b.We have to find the linear speed of its tip at this  angular velocity if the plane is stationary on the tarmac.

v=r\omega=1.15\times 125.7=144.56 m/s

c.Centripetal acceleration=\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2

Centripetal acceleration==\frac{18170.56\times g}{9.81}=1852.25 g m/s^2

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4 years ago
A boat moves with a speed of +2.5 m/s in a direction 25° north of east. If the mass of the boat is 15,000 kg, what is the moment
Marianna [84]

C!!!+37,500 is the answer

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4 years ago
A tight knot can be easily opened by using a long spanner. Give reason.
olga nikolaevna [1]

Answer:

Torque

Explanation:

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You are testing a new roller coaster ride in which a car of mass mm moves around a vertical circle of radius RR. In one test, th
Masja [62]

Answer:

vi = 2.83 √gR

Explanation:

For this exercise we can use the law of conservation of energy

Let's take a reference system that is at point A, the lowest

Starting point. Lower, point A

           Em₀ = Ki = ½ m vi²

Final point. Higher, point B

            Em_{f} = K + U

             

 It indicates that at this point the kinetic energy is ki / 2 and the potential energy is ki / 2

             K = ki / 2

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Energy is conserved so

             Em₀ = Em_{f}

             ½ m vi² = ½ (1/2 m vi²) + m g 2R

             ½ m vi² (1- ½) = m g 2R

              vi² = 4 g 2 R

              vi = √ 8gR = 2 √2gR

              vi = 2.83 √gR

3 0
3 years ago
An object has a mass of 300 g. (a) What is its weight on Earth? (b) What is its mass on the Moon?
Neko [114]

Answer:

The mass of object is 300g.

Its weight on earth is W×0.3kg×9.8m/s2=2.94 N

Its weight on moon is A 300 g would be 48 g on the moon

GOOD LUCK!

7 0
3 years ago
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