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3 years ago

How much energy does a 100 W light bulb use in 24 hours?? Show all your work....

Physics

1 answer:

GalinKa [24]3 years ago

7 0

How much does it use per hour or minute or second? its impossible to answer this question i think

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An isolated conducting sphere has a 15 cm radius. One wire carries a current of 1.0000046 A into it while another wire carries a

lbvjy [14]

Answer: It would take 3.23ms

Explanation: Please see the attachments below

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3 years ago

A particular spring has a spring constant of 25 N/m.

kotegsom [21]

Answer:

$e.p.e = \frac{1}{2} k {x}^{2} \\ = \frac{1}{2} \times 25 \times (0.015m)^{2} \\ = 0.0028125$

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3 years ago

If a guitar string has a fundamental frequency of 500 hz, which one of the following frequencies can set the string into resonan

Sav [38]

I had to look for the options and here is my answer:

Given that a guitar string has a basic frequency of 500hz, the frequency that can be set for this into resonant vibration would be 1500 Hz. The resonant vibration here is necessary in musical instruments because it delivers a vibrating system which leads to a higher amplitude at a particular frequency.

3 0

3 years ago

A rocket is launched straight up from Earth's surface at a speed of 15,000 m/s. What is its speed when it is very far away from

Colt1911 [192]

Answer:

1/2 m v2^2 = 1/2 m v1^2 - G M m / R conservation of energy

v1^2 - v2^2 = 2 G M / R rearranging terms

2 G M / R = 2 * 6.67 * 5.98 / 6.37 E7 = 1.25E8

v1^2 - v2^2 = 1.25E8

v2^2 = v1^2 - 1.25E8 = (1.5^2 - 1.25) * E8

v2 = 1.00E4 = 10,000 m/s

3 0

2 years ago

A 0.140 kg stone rests on a frictionless, horizontal surface. A bullet of mass 8.50 g , traveling horizontally at 320 m/s , stri

Novay_Z [31]

Answer:

a) $v = 34.607\,\frac{m}{s}$ (Positive), b) $e = 0.803$ . The collision is not perfectly elastic.

Explanation:

a) The collision can be described by the Principle of Momentum Conservation and Principle of Energy Conservation:

$(0.140\,kg)\cdot (0\,\frac{m}{s} ) + (0.0085\,kg)\cdot (320\,\frac{m}{s} ) = (0.0085\,kg)\cdot (-250\,\frac{m}{s} ) + (0.140\,kg)\cdot v$

The final velocity of the rock is:

$v = 34.607\,\frac{m}{s}$

b) The coefficient of restitution is the best criterion to distinguish elastic collsions from inelastic collisions, such criterion is the ratio of final energy of the system to initial energy of the system:

$e = \frac{\frac{1}{2}\cdot [(0.140\,kg)\cdot (34.607\,\frac{m}{s} )^{2}+(0.0085\,kg)\cdot (-250\,\frac{m}{s} )^{2}] }{\frac{1}{2}\cdot [(0.140\,kg)\cdot (0\,\frac{m}{s} )^{2}+(0.0085\,kg)\cdot (320\,\frac{m}{s} )^{2}] }$

$e = 0.803$

The collision is not perfectly elastic.

6 0

4 years ago

Read 2 more answers