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valina [46]
3 years ago
13

Bryan Allen pedaled a human-powered aircraft across the English channel from the cliffs of Dover to Cap Gris-Nez on June 12, 197

9. He flew for 169 minutes at an average velocity of 3.53 m/s in a direction 45.0∘ degrees south of east. Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the earth. What was his average velocity relative to the air?
Physics
1 answer:
tigry1 [53]3 years ago
4 0

Answer:

The answer is "5.53 \ \frac{m}{s}"

Explanation:

apply the formula for calculating the average velocity to the relative air

V_{PG} =V_{PA}+V_{AG}

\Rightarrow  V_{PA} = V_{PG} -V_{AG}

Given value:

V_{AG} = -2 \ \frac{m}{s}

V_{PG} =3.53

\Rightarrow  V_{PA} = 3.53 - (-2) \\\\\Rightarrow  V_{PA} = 3.53 +2 \\\\\Rightarrow  V_{PA} = 5.53  \\\\

The final answer is "5.53 \ \frac{m}{s}" in the south-east direction.

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Explain this quote "ambition beats genius 99% of the time. "
klasskru [66]

Answer:

The idea that hard work is the most important aspect of new inventions existed before Edison gave his quote, however.

The idea behind this quote is that it is easy to have a good idea, or a creative insight. However, to follow through with that idea, and turn it into a reality, takes a level of patience and dedication that few people have.

Explanation:

7 0
3 years ago
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A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed
Tresset [83]
For this case we first think that the skateboard and the child are one body.
 We have then:
 1 = jug
 2 = skateboard + boy
 By conservation of the linear amount of movement:
 M1V1i + M2V2i = M1V1f + M2V2f
 Initial rest:
 v1i = v2i = 0
 0 = M1V1f + M2V2f
 Substituting values
 0 = (7.8) (3.2) + (M2) (- 0.65)
 0 = 24.96 + M2 (-0.65)
 -24.96 = (-0.65) M2
 M2 = (-24.96) / (- 0.65) = 38.4 kg
 Then, the child's mass is:
 M2 = Mskateboard + Mb
 Clearing:
 Mb = M2-Mskateboard
 Mb = 38.4 - 1.9
 Mb = 36.5 Kg
 answer:
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4 0
3 years ago
Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10
gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

\Sigma \tau=0

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

-\tau_l+\tau_p+\tau_{cm}=0          (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

\tau=Fd           (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N

The force applied by the left support is 2123.33 N

8 0
3 years ago
The net horizontal force on a car is 981 N. The car has a mass of 1550 kg and the force is applied when the car has a speed of 2
viktelen [127]

Answer:

Distance, d = 778.05 m                          

Explanation:

Given that,

Force acting on the car, F = 981 N

Mass of the car, m = 1550 kg

Initial speed of the car, v = 25 mi/h = 11.17 m/s

We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:

F=ma\\\\a=\dfrac{F}{m}\\\\a=\dfrac{981}{1550}\\\\a=0.632\ m/s^2

Let d is the distance covered by car. Using second equation of motion as :

d=ut+\dfrac{1}{2}at^2\\\\d=11.17\times 35+\dfrac{1}{2}\times 0.632\times (35)^2\\\\d=778.05\ m

So, the car will cover a distance of 778.05 meters.

5 0
4 years ago
Where would a car traveling on a roller coaster have the most kinetic energy ? and why?
goldenfox [79]

Answer:

As the car travels up the coaster it is gaining potential energy.

Explanation:

Because It has the greatest in amount of potential energy at the top of the coaster. when the car travels down the roller coaster it obtains speed and kinetic energy.

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3 years ago
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