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a_sh-v
1 month ago
11

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. After a short rest at the lake, she hi

kes 2.7 km in a direction of 16° east of south to the scenic overlook. What is the magnitude of the hiker’s resultant displacement? Round your answer to the nearest tenth.  km What is the direction of the hiker’s resultant displacement? Round your answer to nearest whole degree. °  south of west
Physics
2 answers:
Softa [3K]1 month ago
6 0
The hiker’s overall displacement magnitude measures 5.6 km.
<span>The direction of this displacement is 77 degrees</span> south of west.
Maru [3.3K]1 month ago
5 0

1) Magnitude

Assigning south as the positive y-axis and east as the positive x-axis, we break down the two displacements into components:

d_{1x} = -(3.5 km) cos 55^{\circ}=-2.0 km

d_{1y} = (3.5 km) sin 55^{\circ}=2.87 km

d_{2x} = (2.7 km) sin 16^{\circ}=0.74 km

d_{2y} = (2.7 km) cos 16^{\circ}=2.60 km

The total displacement components resolve to:

d_x = d_{1x}+d_{2x}=-2.0 km +0.74 km=-1.26 km east (equivalent to 1.26 km west)

d_y=d_{1y}+d_{2y}=2.87 km + 2.60 km=5.47 km south

Calculating the resultant displacement's magnitude yields:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(1.26)^2+(5.47)^2}=5.61 km


2) Direction

The displacement's direction is found to be:

\theta= arctan(\frac{d_y}{d_x})=arctan(\frac{5.47}{1.26})=arctan(4.34)=77.0^{\circ} south of west.

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Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g
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Answer:

Q = 12.5 kJ

Explanation:

The formula used to compute heat is:

Q = H° * n

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Q: heat (J or kJ)

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n: moles

Now, as noted in the comments, the question lacks completeness, here is the part that is missing:

Given:

2A + B  A2B (1)

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Using the above two reactions, we need to establish the overall reaction (the one presented in the question), so let’s combine (1) and (2):

2A + B -------> A2B   H°1 = -25 kJ/mol

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When we add these equations, one A2B cancels out with one A2B from reaction 2, thus, we have:

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(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe
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Answer:

(a) the coefficient of friction is 0.451

This was derived using the energy conservation principle (the total energy in a closed system remains constant).

(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.

Explanation:

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Answer:

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