This is a projectile motion problem, so, we use the formula for trajectory:
y =xtanα + gx^2/2v^2(cosα)^2
where
y is the vertical distance (y = 50 m)
x is the horizontal distance (x=90 m)
α is the angle of trajectory; since it levels of HORIZONTALLY, α = 0°
v is the initial velocity
g is the acceleration due to gravity which is 9.81 m/s^2
Substituting to the formula,
50 =90tan(0°) + (9.81)(90)^2/2v^2(cos0°)^2
v = 28.2 m/s
Answer:
Explanation:
To solve this problem we have to take into account that the energy consumed per second is the power. Hence, by multipling the power and the time spent to arrive to the lab we obtain the total energy consumed.
But first we have to calculate the time
Now we use E=W*t for both times
A. Hence, by running the energy consumed is lower.
B.
E1=1008000J
E2=1392000J
C. Because the more intense exercise is made in a lower time in comparison with the less intense exercise, and higher the time, more energy is consumed.
Answer: 61.1 is the answer I wish this answer help you.
<em>optical</em><em> </em><em>fibre</em>
The distance covered in one orbit is equal to the circumference of earth.
Given that radius of earth is 3959 miles.
Circumference of earth= 2πr = 2*3.14*3959 = 24875.13 miles
Speed is 26,000 miles per hour.
Since, Velocity= Distance/Time
Time= Distance/Velocity = 24875.13/26000
Time = 0.9567 hours
Therefore,
Average hour per orbit is 0.957 approximately.