The answer:
the full question is as follow:
<span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D?
As shown in the figure,
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
D = 0 - (A + B + C) ,
let W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics
Magnitude
A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
the direction of D can be found by using Dx and Dy value
we know that tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°
direction is 26.02°
Answer:
A. 2.36
Explanation:
This is the right answer for Plato users!
Answer:
It is false, many fossils have bones and fur (etc) in them
Answer:
e) half
Explanation:
as we know relation b/w charge(Q) , capacitance (C) and voltage (V) is
Q = CV ...........(i)
we also know capacitance of a capacitor for separation d and area A is
C = ∈₀A/d ...........(ii)
from the above equation it is clear that C is inversely proportional to separation if separation doubled capacitance will reduced to half and from equation first charge is directly proportional to capacitance.
that s why magnitude of charge will reduced to half
so option (e) is right answer
Answer:
-209.42J
Explanation:
Here is the complete question.
A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?
Solution
The work done by a force W = ∫Fdx since our force is variable.
Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.
So, the workdone by the force on the cow is
W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx
= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx
= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹
= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]
= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J
= -209.415 J ≅ -209.42J