The figure displays an overhead view of a 0.023 kg lemon half, along with two of the three horizontal forces acting on it while on a frictionless table.- Force \(\vec{F}_1\) has a magnitude of 7 N and is at an angle \(\theta_1 = 29^\circ\).- Force \(\vec{F}_2\) has a magnitude of 11 N and is at an angle \(\theta_2 = 25^\circ\).In unit-vector notation, determine the third force if the lemon half:- (a) is stationary,- (b) has a constant velocity \(\vec{v} = (10\hat{i} - 15\hat{j})\) m/s, and- (c) has an acceleration \(\vec{v} = (10t\hat{i} - 12t\hat{j})\) m/s\(^2\), where \(t\) is time.**Diagram Explanation:**The diagram shows:- A lemon half at the origin.- Force \(\vec{F}_1\) positioned in the first quadrant at an angle \(\theta_1 = 29^\circ\) from the negative \(y\)-axis.- Force \(\vec{F}_2\) positioned in the fourth quadrant at an angle \(\theta_2 = 25^\circ\) from the positive \(x\)-axis.**Inputs Required:**(a) For stationary, input values for the third force in unit-vector form.(b) For constant velocity, input values for the third force in unit-vector form.(c) For specified acceleration, input values for the third force in unit-vector form. Each scenario requires values for \(\hat{i}\) and \(\hat{j}\) components along with their units.

Answered: Image /qna-images/answer/de407a2f-6125-4fa0-b6f8-6c903fe7c352.jpg

1 The figure shows an overhead view of a 0.023 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force ₁ has a magnitude of 7 N and is at 0₁ = 29°. Force F2 has a magnitude of 11 N and is at 0₂ = 25°. In unit-vector notation, what is the third force if the lemon half (a) is stationary, (b) has the constant velocity = (10î – 15ĵ) m/s, and (c) has the ✓ = (10fî – 12ţĵ) m/s², where t is time? (a) Number i (b) Number i (c) Number i F 0₁ i + i i + i i + i 0₂ AF₂ B j Units j Units j Units < <

1 The figure shows an overhead view of a 0.023 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force ₁ has a magnitude of 7 N and is at 0₁ = 29°. Force F2 has a magnitude of 11 N and is at 0₂ = 25°. In unit-vector notation, what is the third force if the lemon half (a) is stationary, (b) has the constant velocity = (10î – 15ĵ) m/s, and (c) has the ✓ = (10fî – 12ţĵ) m/s², where t is time? (a) Number i (b) Number i (c) Number i F 0₁ i + i i + i i + i 0₂ AF₂ B j Units j Units j Units < <

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

The figure displays an overhead view of a 0.023 kg lemon half, along with two of the three horizontal forces acting on it while on a frictionless table.

- Force \(\vec{F}_1\) has a magnitude of 7 N and is at an angle \(\theta_1 = 29^\circ\).
- Force \(\vec{F}_2\) has a magnitude of 11 N and is at an angle \(\theta_2 = 25^\circ\).

In unit-vector notation, determine the third force if the lemon half:
- (a) is stationary,
- (b) has a constant velocity \(\vec{v} = (10\hat{i} - 15\hat{j})\) m/s, and
- (c) has an acceleration \(\vec{v} = (10t\hat{i} - 12t\hat{j})\) m/s\(^2\), where \(t\) is time.

**Diagram Explanation:**

The diagram shows:
- A lemon half at the origin.
- Force \(\vec{F}_1\) positioned in the first quadrant at an angle \(\theta_1 = 29^\circ\) from the negative \(y\)-axis.
- Force \(\vec{F}_2\) positioned in the fourth quadrant at an angle \(\theta_2 = 25^\circ\) from the positive \(x\)-axis.

**Inputs Required:**

(a) For stationary, input values for the third force in unit-vector form.

(b) For constant velocity, input values for the third force in unit-vector form.

(c) For specified acceleration, input values for the third force in unit-vector form.

Each scenario requires values for \(\hat{i}\) and \(\hat{j}\) components along with their units.

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