The missing values in the table for n =1, 2, 3, 4 and 5.

Question

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Answer 1

Question

Chapter 26, Problem 78QAP

To determine

The missing values in the table for n =1, 2, 3, 4 and 5.

Expert Solution & Answer

Answer to Problem 78QAP

The final table is,

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Explanation of Solution

Given info:

The given table is,

n	L_n	r_n	K_n	E_n	v_n
n= 1
n= 2
n= 3
n= 4
n= 5

Formula used:

The expressions for angular momentum, speed, orbital radius, kinetic energy and total energy are:

Ln=nh2πrn=n2h24π2mke2Kn=mk2e42n2ℏ2En=−mk2e42n2ℏ2vn=2πke2nh

Calculation:

All values are given in SI unit. The values of different constants as considered for calculation are
k=8.99×109 N⋅m2⋅Coulomb-2m=9.11×10−31 kgh =1.055×10-34 Joule-sece =1.6×10-19 Coulomb

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

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The force of the quadriceps (Fq) and force of the patellar tendon (Fp) is identical (i.e., 1000 N each). In the figure below angle in blue is Θ and the in green is half Θ (i.e., Θ/2). A) Calculate the patellar reaction force (i.e., R resultant vector is the sum of the horizontal component of the quadriceps and patellar tendon force) at the following joint angles: you need to provide a diagram showing the vector and its components for each part. a1) Θ = 160 degrees, a2) Θ = 90 degrees. NOTE: USE ONLY TRIGNOMETRIC FUNCTIONS (SIN/TAN/COS, NO LAW OF COSINES, NO COMPLICATED ALGEBRAIC EQUATIONS OR ANYTHING ELSE, ETC. Question A has 2 parts!

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Answer 2

Question

Chapter 26, Problem 78QAP

To determine

The missing values in the table for n =1, 2, 3, 4 and 5.

Expert Solution & Answer

Answer to Problem 78QAP

The final table is,

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Explanation of Solution

Given info:

The given table is,

n	L_n	r_n	K_n	E_n	v_n
n= 1
n= 2
n= 3
n= 4
n= 5

Formula used:

The expressions for angular momentum, speed, orbital radius, kinetic energy and total energy are:

Ln=nh2πrn=n2h24π2mke2Kn=mk2e42n2ℏ2En=−mk2e42n2ℏ2vn=2πke2nh

Calculation:

All values are given in SI unit. The values of different constants as considered for calculation are
k=8.99×109 N⋅m2⋅Coulomb-2m=9.11×10−31 kgh =1.055×10-34 Joule-sece =1.6×10-19 Coulomb

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

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Answer 3

Question

Chapter 26, Problem 78QAP

To determine

The missing values in the table for n =1, 2, 3, 4 and 5.

Expert Solution & Answer

Answer to Problem 78QAP

The final table is,

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Explanation of Solution

Given info:

The given table is,

n	L_n	r_n	K_n	E_n	v_n
n= 1
n= 2
n= 3
n= 4
n= 5

Formula used:

The expressions for angular momentum, speed, orbital radius, kinetic energy and total energy are:

Ln=nh2πrn=n2h24π2mke2Kn=mk2e42n2ℏ2En=−mk2e42n2ℏ2vn=2πke2nh

Calculation:

All values are given in SI unit. The values of different constants as considered for calculation are
k=8.99×109 N⋅m2⋅Coulomb-2m=9.11×10−31 kgh =1.055×10-34 Joule-sece =1.6×10-19 Coulomb

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

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Answer 4

Question

Chapter 26, Problem 78QAP

To determine

The missing values in the table for n =1, 2, 3, 4 and 5.

Expert Solution & Answer

Answer to Problem 78QAP

The final table is,

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Explanation of Solution

Given info:

The given table is,

n	L_n	r_n	K_n	E_n	v_n
n= 1
n= 2
n= 3
n= 4
n= 5

Formula used:

The expressions for angular momentum, speed, orbital radius, kinetic energy and total energy are:

Ln=nh2πrn=n2h24π2mke2Kn=mk2e42n2ℏ2En=−mk2e42n2ℏ2vn=2πke2nh

Calculation:

All values are given in SI unit. The values of different constants as considered for calculation are
k=8.99×109 N⋅m2⋅Coulomb-2m=9.11×10−31 kgh =1.055×10-34 Joule-sece =1.6×10-19 Coulomb

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

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Answer 5

Chapter 26, Problem 78QAP

To determine

The missing values in the table for n =1, 2, 3, 4 and 5.

Expert Solution & Answer

Answer to Problem 78QAP

The final table is,

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Explanation of Solution

Given info:

The given table is,

n	L_n	r_n	K_n	E_n	v_n
n= 1
n= 2
n= 3
n= 4
n= 5

Formula used:

The expressions for angular momentum, speed, orbital radius, kinetic energy and total energy are:

Ln=nh2πrn=n2h24π2mke2Kn=mk2e42n2ℏ2En=−mk2e42n2ℏ2vn=2πke2nh

Calculation:

All values are given in SI unit. The values of different constants as considered for calculation are
k=8.99×109 N⋅m2⋅Coulomb-2m=9.11×10−31 kgh =1.055×10-34 Joule-sece =1.6×10-19 Coulomb

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Answer 6

Chapter 26, Problem 78QAP

To determine

The missing values in the table for n =1, 2, 3, 4 and 5.

Expert Solution & Answer

Answer to Problem 78QAP

The final table is,

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Explanation of Solution

Given info:

The given table is,

n	L_n	r_n	K_n	E_n	v_n
n= 1
n= 2
n= 3
n= 4
n= 5

Formula used:

The expressions for angular momentum, speed, orbital radius, kinetic energy and total energy are:

Ln=nh2πrn=n2h24π2mke2Kn=mk2e42n2ℏ2En=−mk2e42n2ℏ2vn=2πke2nh

Calculation:

All values are given in SI unit. The values of different constants as considered for calculation are
k=8.99×109 N⋅m2⋅Coulomb-2m=9.11×10−31 kgh =1.055×10-34 Joule-sece =1.6×10-19 Coulomb

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Answer 7

Chapter 26, Problem 78QAP

To determine

The missing values in the table for n =1, 2, 3, 4 and 5.

Answer 8

Expert Solution & Answer

Answer to Problem 78QAP

The final table is,

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given info:

The given table is,

n	L_n	r_n	K_n	E_n	v_n
n= 1
n= 2
n= 3
n= 4
n= 5

Formula used:

The expressions for angular momentum, speed, orbital radius, kinetic energy and total energy are:

Ln=nh2πrn=n2h24π2mke2Kn=mk2e42n2ℏ2En=−mk2e42n2ℏ2vn=2πke2nh

Calculation:

All values are given in SI unit. The values of different constants as considered for calculation are
k=8.99×109 N⋅m2⋅Coulomb-2m=9.11×10−31 kgh =1.055×10-34 Joule-sece =1.6×10-19 Coulomb

n	L_n	r_n	K_n	E_n	v_n
	Ln=nh2π Kg⋅meter2⋅s-1	rn=n2h24π2mke2 meter	Kn=mk2e42n2ℏ2 Joule	En=−mk2e42n2ℏ2 Joule	vn=2πke2nh meter⋅s-1
n= 1	1.06×10−34	0.536×10−10	21.44×10−19	−21.44×10−19	21.47×105
n= 2	2.12×10−34	2.144× 10 −10	5.36×10−19	−5.36×10−19	10.73×105
n= 3	3.18×10−34	4.824×10−10	2.38×10−19	−2.38×10−19	7.16×105
n= 4	4.24×10−34	8.576×10−10	1.34×10−19	−1.34×10−19	5.36×105
n= 5	5.3×10−34	13.4×10−10	0.85×10−19	−0.85×10−19	4.29×105

Answer 12

Ch. 26 - Prob. 1QAP Ch. 26 - Prob. 2QAP Ch. 26 - Prob. 3QAP Ch. 26 - Prob. 4QAP Ch. 26 - Prob. 5QAP Ch. 26 - Prob. 6QAP Ch. 26 - Prob. 7QAP Ch. 26 - Prob. 8QAP Ch. 26 - Prob. 9QAP Ch. 26 - Prob. 10QAP

The missing values in the table for n =1, 2, 3, 4 and 5.

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The missing values in the table for n =1, 2, 3, 4 and 5.

Answer to Problem 78QAP

Explanation of Solution

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Chapter 26 Solutions