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prohojiy [21]
3 years ago
7

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i

s performed in three stages, each requiring a vertical distance of 12.0 m: (a) the initially stationary spelunker is accelerated to a speed of 2.90 m/s; (b) he is then lifted at the constant speed of 2.90 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 77.0 kg rescue by the force lifting him during each stage
Physics
1 answer:
Leya [2.2K]3 years ago
5 0

Answer:

a) 323.4J

b) 0J

c) -323.4J

Explanation:

a) W=Fd

F=ma

solve for acc. using kinematics

v^2=vo^2+2a(x)

8.41=2a(12)

4.205=a(12)

0.35=a

F=(77)(0.35)

F=26.95N

W=26.95*12...... W=323.4J

b) No acceleration, thus no force, thus no work!

c) W=Fd

F=ma

find acc. using kinematics: v^2=vo^2+2a(x)

0=(2.9^2)+2a(12)

0=8.41+2a(12)

-8.41=2a(12)

-4.205=a(12)

-0.35=a

F=(77)(-0.35)

F=-26.95N

W=(-26.95)(12)

W=-323.4J

Yes, work can be negative!

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R=∆V/I=12/0.6=20ohm....

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Answer:

Y, X, Z, W

Explanation:

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Calculate the electric field at point A, located at coordinates (0 m, 12.0 m ). Give the x and y components of the electric fiel
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Answer:

The correct answer is "(0,300\times 10^{-3} \ N/C)".

Explanation:

The given problem seems to be incomplete. Please find the attachment of the complete query.

According to the question,

At point A, we have

⇒ E_x = \frac{k q_1}{d_1^2} Cos \theta_1 -   \frac{k q_2}{d_2^2} Cos \theta_2

or,

⇒ E_x = 9\times 10^9\times [\frac{6\times 10^{-9}}{15^2}\times \frac{9}{15}-\frac{8\times 10^{-9}}{20^2}\times \frac{16}{20} ]

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and,

⇒ E_y = \frac{kq_1}{d_1^2}Sin \theta_1 +\frac{kq_2}{d_2^2}Sin \theta_2

or,

⇒ E_y = 9\times 10^9\times [\frac{6\times 10^{-9}\times 12}{15^2\times 15}+ \frac{8\times 10^{-9}\times 12}{20^2\times 20} ]

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Two identical metal balls of radii 2.50
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Answer:

+1.33 × 10^{-7} C and -1.33 × 10^{-7} C respectively.

Explanation:

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where: k is a constant = 9 × 10^{9} Nm^{2} C^{-2}, q is the charge and r is the distance between the charges.

From equation 1,

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The charge on each ball can be determined as;

given that; V = 1.2 × 10^{3}, k = 9 × 10^{9} Nm^{2} C^{-2} and r = 1.00 m.

From equation 2,

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Thus, the charge on the first ball is +1.33 × 10^{-7} C, while the charge on the second ball is -1.33 × 10^{-7} C.

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