Answer:
A. 0.044 kg
Explanation:
We need to subtract the sum of (beaker+water - empty beaker) which is 0.106 kg - 0.062 kg = 0.044 kg. The answer will not be written in Newton because this unit is used for force only and in this question w have to find the weight.
Hope it is enough.
Please mark me as brainliest.
Answer:
421.83 m.
Explanation:
The following data were obtained from the question:
Height (h) = 396.9 m
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
First, we shall determine the time taken for the ball to get to the ground.
This can be calculated by doing the following:
t = √(2h/g)
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 396.9 m
Time (t) =.?
t = √(2h/g)
t = √(2 x 396.9 / 9.8)
t = √81
t = 9 secs.
Therefore, it took 9 secs fir the ball to get to the ground.
Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:
Time (t) = 9 secs.
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
s = ut
s = 46.87 x 9
s = 421.83 m
Therefore, the horizontal distance travelled by the ball is 421.83 m
Answer:
D
Explanation:
If you list and compare, the elephants hearing range is the closest to a humans.
Humans: 20-20,000
Elephant: 16-12,000
While the others are either too much more or less than a humans.
Bat: 2,000-110,000
Dolphin: 90-105,000
Chicken: 125-2,000
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
