Answer:
Explanation:
We shall represent all the forces in vector form .
Force of 95 pounds
F₁ = 95cos100 i + 95sin100j
= -16.5 i +93.55 j
force of 75 pounds
F₂ = 75cos200 + 75sin200j
= -70.47 i - 25.65 j
force of 146 pounds
F₃ = 146cos300 i + 146sin300j
= 73i -126.44 j
Resultant force
R = F₁+ F₂ + F₃
= -16.5 i +93.55 j -70.47 i - 25.65 j +73i -126.44 j
= -13.97 i - 58.54 j
Magnitude of R = √ ( 13.97² + 58.54² )
= 60.18
If Ф be angle of resultant with axis
tanФ = - 58.54 / -13.97
Ф = 76.57 + 180 = 256.57 , because both x and y components are negative. So the resultant will be in third quadrant.
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Answer:
B
Explanation:
In this calorimetry problem, the heat released by the reaction is equal to the heat absorbed by the solution (assumed to have the same specific heat capacity as water, 4.19 Jg⁻¹°C⁻¹).
The formula Q = mcΔt will be used to calculate the heat energy, where m is the mass, c is the specific heat capacity, and Δt is the change in temperature from final to initial.
The volume of solution is (50.0 + 50.0)mL = 100.0mL = 100.0g, since water has a density of 1.00g/mL.
The heat absorbed by the solution is then calculated.
Q = mcΔt = (100.0 g)(4.19 Jg⁻¹°C⁻¹)(28.2°C - 25.0°C) = 1340 J
The closest answer is B) 1300 J. This answer is obtained by including only two significant figures in the answer.
<span>Integrate a to get v and use initial data for v to evaluate the constants of integration.
v = [(3/2)t^2 + Ci]i + [2t^2 + Cj]j
When t=0, v = 5i + 2j, hence Ci=5, Cj=2
So v = [(3/2)t^2 + 5]i + [2t^2 + 2]j <==ANS
(b)
Integrate v to get r and use initial data for r to evaluate the constants of integration:
The result is:
r = [0.5t^3 + 5t + 20]i + [(2/3)t^3 + 2t + 40]j <==ANS
Set t=4 in the above expression to evaluate (c)
Convert the result of (c) to polar form to evaluate (d)</span>
Answer:
no its D prep said it wasnt a
Explanation:
IT IS D NOT A
