The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
To find the answer, we need to know about the Newton's equation of motion.
<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>
- The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
- V= final velocity, U= initial velocity, S= distance and a= acceleration
<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>
- Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
- So, 100²-0= 2×a×1800
=> 10000= 3600a
=> a= 10000/3600 = 2.8 m/s²
Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
Learn more about the Newton's equation of motion here:
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Answer:
After refraction at two parallel faces of a glass slab, a ray of light emerges in a direction parallel to the direction of incidence of white light on the slab. As rays of all colours emerge in the same direction (of incidence of white light), hence there is no dispersion, but only lateral displacement.
<span>3. The attempt at a solution So basically what I did was divided into components. x: (3)(2000) = (3000)*v_x y: (v_vw)*(10000) = (3000)*v_y v_x, v_y is the velocity (after collision) in the x and y direction, respectively, of both cars stuck together (since it is an inelastic collision). v_vw is the initial velocity of the Volkswagen. Now what I did was that the angle is 35 degrees north of east. So basically made a triangle and figured that tan(35) = (v_y)/(v_x). This means (v_x)*(tan35) = v_y. Then, I simplified the component equations to get: x: 2 = v_x y: v_vw = 3*v_y Then plugging in for v_y, I got: v_vw = 3(2)(tan35) = 4.2 m/s as the velocity of the volkswagen. However, the answer key says 8.6 m/s. Could someone please help me out? Thanks Phys.org - latest science and technology news stories on Phys.org • Game over? Computer beats human champ in ancient Chinese game • Simplifying solar cells with a new mix of materials • Imaged 'jets' reveal cerium's post-shock inner strength Oct 24, 2012 #2 ehild Homework Helper Gold Member What directions you call x and y?
Reference https://www.physicsforums.com/threads/2d-momentum-problem.646613/</span>
Answer:
1.5 m/s²
Explanation:
Applying,
a = (v-u)/t ...................... Equation 1
Where a = acceleration of the particle, v = final velocity of the particle, u = Initial velocity of the particle, t = time.
From the question,
Given: u = 0 m/s (From rest), v = 15 m/s, t = 10 seconds
Substitute these values into equation 1
a = (15-0)/10
a = 15/10
a = 1.5 m/s²
