The average acceleration from 9 to 18 seconds is 6 meters per second.
The graph shows that from 9 to 18 seconds the speed of the ostrich stays at a steady 6 meters per second.
<u>The possible formulas for impulse are as follows:</u>
J = FΔt
J = mΔv
J = Δp
Answer: Option A, E and F
<u>Explanation:</u>
The quantity which explains the consequences of a overall force acting on an object (moving force) is known as impulse. It is symbolised as J. When the average overall force acting on an object than such products are formed and in given duration than the start fraction force over change in time end fraction J = FΔt.
The impulse-momentum theorem explains that the variation in momentum of an object is same as the impulse applied to it: J = Δp J = mΔv if mass is constant J = m dv + v dm if mass changes. Logically, the impulse-momentum theorem is equivalent to Newton second laws of motion which is also called as force law.
Answer:
Why are continental rises and abyssal plains relatively rare in the Pacific? This is because the extensive system of trenches along the active margins of the Pacific, trap much of the sediments flowing off the continents, preventing them from building the broad, flat abyssal plains typical of the Atlantic ocean basins.
Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
![E = \frac{T}{P*sin(\theta)}= \frac{7.3 X 10^{-9}}{1.476X10^{-11}*sin(36.8)}\\\\E = 825.6 N/C](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BT%7D%7BP%2Asin%28%5Ctheta%29%7D%3D%20%20%5Cfrac%7B7.3%20X%2010%5E%7B-9%7D%7D%7B1.476X10%5E%7B-11%7D%2Asin%2836.8%29%7D%5C%5C%5C%5CE%20%3D%20825.6%20N%2FC)
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C