Answer:
momentum=mass×velocity
momentum =400kg×20m/s=8000kg.m/s
<span>9.50x10^2 newtons
A pascal is defined as 1 newton per square meter. So let's multiply the pressure by the surface area of the box lid.
F = 1.00x10^-2 m^2 * 9.50x10^4 N/m^2 = 9.50x10^2 N
So it will take 9.50x10^2 newtons of force to remove the lid from the box.</span>
Explanation:
Let us assume that the maximum allowable horizontal distance be represented by "d".
Therefore, torque equation about A will be as follows.

d = ![\frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20%5Ctimes%2075%20%5Ctimes%20%280.7%2B0.15%2B0.15%29%20-%2060%20%5Ctimes%200.15%20-%20252%20%5Ctimes%200.15%20%5Ctimes%202%5D%7D%7B252%7D)
d = 0.409 m
Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.
Explanation:
first you have to find accelerarion, it is given that the initial velocity(u) is 3 m/s, distance travelled(s) be 2m finall it came to rest so final velocity be 0m/s
now using the 3rd law of motion
v^2=u^2+2as
0=9+2a2
a= -9/4m/s^2
now force=mass×accelration
=2kg×(-9/4)m/s^2
=4.5 N
4.5 newton force applied on the book!
✌️:)
Answer:
Work done = 4584.9 J
Explanation:
given: q1=3.0 mC = 3.0 × 10⁻³ C, r = 20 cm = 0.20 m, q1 = 34μC = 34 × 10⁻⁶ C
Solution:
Formula for the potential difference at the center of the circle
P.E = K × q1 q2 /r (Coulomb's constant k= 8.99 × 10⁹ N·m² / C²)
P.E = 8.99 × 10⁹ N·m² / C² × 3.0 × 10⁻³ C × 34 × 10⁻⁶ C / 0.20 m
P.E = 4584.9 J = Work done