Answer:
11.625
Explanation:
L = length of the ladder = 16 ft
= rate at which top of ladder slides down = - 3 ft/s
= rate at which bottom of ladder slides
y = distance of the top of ladder from the ground
x = distance of bottom of ladder from wall = 4 ft
Using Pythagorean theorem
L² = x² + y²
16² = 4² + y²
y = 15.5 ft
Also using Pythagorean theorem
L² = x² + y²
Taking derivative both side relative to "t"



= 11.625 ft/s
Answer:
please the answer below
Explanation:
(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

k=8.89*10^9
For both cases we have

(b) by replacing this values of r in the expression for V we obtain

hope this helps!!
Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:

where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.
We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force

acting in the opposite direction. So Newton's second law can be rewritten as

or

since the frictional force is 15 N and we want to achieve an acceleration of

, we can substitute these values to find what is the force the man needs:
Answer:
ΔE = 37.8 x 10^9 J
Explanation:
The energy required will increased the potential energy and increase the kinetic energy.
As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative
Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²
G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²
ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J
The centripetal force at orbit must be equal to the gravity force
mv²/R = mg'
v²/8.0e6 = 6.272
v² = (6.272(8.0e6)) = 50.2e6 m²/s²
The maximum velocity when resting on earth at the equator is about 460 m/s.
The change in kinetic energy is
ΔKE = ½m(vf² - vi²)(1000)
ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J
Total energy increase is
25e9 + 12.857e9 = 37.8e9 J
Answer:
The direction will be NE i.e North east.
Explanation:
From the question, it will be travelling in North east direction because the the negative X components in vector is in horizontal direction which is the Eastern direction when using the four cardinal points and the Y components, the positive is in North direction if you use the four cardinal points. Therefore, the vector with both X components and Y components will be travelling in Northeast direction.