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cestrela7 [59]
2 years ago
14

A submerged submarine alters its buoyancy so that it initially accelerates upward at 0.325 m/s^2. What is the submarine's averag

e density at this time? (Hint: the density of sea water is 1.025x10^3 kg/m^3.)
Physics
1 answer:
scoundrel [369]2 years ago
7 0
<span>You are given a submerged submarine accelerating upward at 0.325 m/s</span>² and the density of sea water is 1.025x10³ kg/m³. The submarine's average density at this time is 22 kg/m³.
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After an incandescent lamp is turned on, the temperature of its filament rapidly increases from room temperature to its operatin
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Answer: (1) The resistance increases and the current decreases.

Explanation:  

When the temperature of the filament increases, the vibrational energy of the constituent atoms increases which leads to increase in inter-atomic collision. Thus, the resistance would increase. The increases in resistance would obstruct the flow of charges more leading to decrease in the value of the current.

Hence, when the temperature of the filament increase, the resistance increases and current decreases.

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2 years ago
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A pillow is thrown downward with an initial speed of 6 m/s.
Yuri [45]

Given :

Initial velocity, u = -6 m/s.

Time taken, t = 4 seconds.

Acceleration due to gravity, g = -9.8\ m/s^2.( Here negative sign means downward direction )

To Find :

Velocity after 4 seconds.

Solution :

By equation of motion.

v = u + at

Here , a = g.

v = u + gt

v = -6 + (-9.8)×4

v = -6 + (-39.2)

v = -45.2 m/s

Therefore, velocity after 4 seconds is -45.2 m/s.

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2 years ago
Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)
Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force,(FS_p)s=-k\Delta s-q(\Delta s)^3

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q=750 N/m^3

We have to find the work must you do to compress this spring 15 cm.

\Delta s=15 cm=0.15 m

Using 1 m=100 cm

Work done=\int_{0}^{0.15}-Fd(\Delta s)

W=-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)

W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}

W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750

W=4.033 J

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Percentage increase in work=\frac{4.033-3.938}{3.928}\times 100=2.4%

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2 years ago
An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign
saul85 [17]

Answer:

2.1\times 10^{-9} T

Explanation:

We are given that

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1kHz=10^{3} Hz

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1 km=1000 m

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B=\frac{0.63}{3\times 10^8}=0.21\times 10^{-8} T

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olga nikolaevna [1]

Answer:

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