<h3>
Answer:</h3>
225 meters
<h3>
Explanation:</h3>
Acceleration is the rate of change in velocity of an object in motion.
In our case we are given;
Acceleration, a = 2.0 m/s²
Time, t = 15 s
We are required to find the length of the slope;
Assuming the student started at rest, then the initial velocity, V₀ is Zero.
<h3>Step 1: Calculate the final velocity, Vf</h3>
Using the equation of linear motion;
Vf = V₀ + at
Therefore;
Vf = 0 + (2 × 15)
= 30 m/s
Thus, the final velocity of the student is 30 m/s
<h3>Step 2: Calculate the length (displacement) of the slope </h3>
Using the other equation of linear motion;
S = 0.5 at + V₀t
We can calculate the length, S of the slope
That is;
S = (0.5 × 2 × 15² ) - (0 × 15)
= 225 m
Therefore, the length of the slope is 225 m
<span>The moon is smaller and more dense than the Earth, and has less extreme temperature changes. The statement presented is True. In terms of temperature, since there is no atmosphere on the moon, then it has less extreme temperature changes. The moon can reach 253 Fahrenheit in the day and -387 Fahrenheit at night.</span>
Answer:
10m
Explanation:
The object distance and image distance is the same from the mirror. so the image is 5m behind the mirror.
5+5=10
Answer:
The tube should be held vertically and perpendicular to the ground.
Explanation:
Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:
Reasoning:
The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.
Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.
So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.
hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.
Explanation:
Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.
→ The tangential component of acceleration is rate of increase in the speed of plane so,
→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).
dy/dx = d(0.4x²)/dx
= 0.8x
Take the derivative again,
d²y/dx² = d(0.8x)/dx
= 0.8
at x= 5 Km
dy/dx = 0.8(5)
= 4
![p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B%5B1%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%20%7D%7B%5Cfrac%7Bd%5E%7B2y%7D%20%7D%7Bdx%5E%7B2%7D%20%7D%20%7D)
now insert the values,
![p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B%5B1%2B%284%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%7D%7B0.8%7D%20%20%3D%2087.62%20km)
→ Now the normal component of acceleration is given by
= (200)²/(87.6×10³)
aₙ = 0.457 m/s²
→ Now the total acceleration is,
![a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}](https://tex.z-dn.net/?f=a%20%3D%20%5B%28a_%7Bt%7D%29%5E%7B2%7D%20%2B%28a_%7Bn%7D%20%29%5E%7B2%7D%20%5D%5E%7B0.5%7D)
![a = [(0.8)^{2} + (0.457)^{2}]^{0.5}](https://tex.z-dn.net/?f=a%20%3D%20%5B%280.8%29%5E%7B2%7D%20%2B%20%280.457%29%5E%7B2%7D%5D%5E%7B0.5%7D)
a = 0.921 m/s²
