Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).
The answer is 36 kg m/s
As it is given that the air bag deploy in time
total distance moved by the front face of the bag
Now we will use kinematics to find the acceleration
now as we know that
so we have
so the acceleration is 400g for the front surface of balloon
Answer:
I believe the answer is esophagus
Answer:
![\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
Explanation:
The electric field created by an infinitely long wire can be found by Gauss' Law.
For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.
![\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cvec%7BE%7D_1%20%2B%20%5Cvec%7BE%7D_2%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20%2B%20%5Cfrac%7B-%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
