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kaheart [24]
3 years ago
5

A light bulb does 100 joules of work in 2.5 seconds. how much work power dos it have

Physics
1 answer:
Lesechka [4]3 years ago
5 0
100/2.5 because power=energy/time
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It is the final seconds of an ice hockey game between the Flyers and the Bruins. The Bruins are down by 1 point. With 20 s left
liberstina [14]

Answer:

a

P =0.1094 \ m past the goal post

Yes the Flyers won the match

Explanation:

Generally the distance covered by the pluck during the last 1.25 seconds is mathematically represented as

     D =  ut + \frac{1}{2} * a * t^2

=>  D =30  * 1.25 + \frac{1}{2} * (-0.5) * (1.25)^2

=>   D =37.1094 \ m

Generally the position of the puck when the game clock reaches zero is mathematically represented as

     P = 37.1094 -37

       P =0.1094 \ m past the goal post

Given that the Bruins where one point down and Flyers scored another goal  it means that the flyer are now two point up hence they won the match

6 0
3 years ago
A voltage of 21.0V across the terminals of a device produces a current of 3.00 mA through the device. What is the resistance of
densk [106]

Answer: 7,000 ohms

Explanation:

From the equation of ohms law V= IR

V=21 volt

I = 3 MA = 3.0×10^-3 A

So by evaluating, we have.

V= IR

21= 3.0×10^-3 × R

Then R = 21/3.0×10^-3

R= 7,000 ohms//.

8 0
3 years ago
Electromagnets are created by
VladimirAG [237]

Answer:

electromagnet: A magnet made of an insulated wire coiled around an iron core (or any magnetic material such as iron, steel, nickel, cobalt) with electric current flowing through it to produce magnetism. The electric current magnetizes the core material.

Explanation:

4 0
3 years ago
Read 2 more answers
If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to
-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm

Moment of inertia from neutral axis will be given by \frac {bd^{3}}{12}+ ay^{2}

Therefore, moment of inertia will be

\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}

Bending stress at top= \frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa

Bending stress at bottom=\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03 Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

8 0
3 years ago
Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. Ass
Naddika [18.5K]

Answer:

L = 4.711 *10^{-6} kg m2/s

Explanation:

i =\frac{ml^3}{3}

= \frac{0.00*.15^2}{3}

   =4.5*10^-5

angular velocity

\omega = \frac{2\pi}{60}

             = 0.1047 rad/s

the angular momentum,

L = I\omega

L = 4.5*10^{-5}* 0.1047 rad/s

L = 4.711 *10^{-6} kg m2/s

3 0
2 years ago
Read 2 more answers
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