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3 years ago

Scientist measure weight in what

Physics

1 answer:

stealth61 [152]3 years ago

3 0

kilograms or mass hope this helps

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A sinusoidal traveling wave is generated on a string by an oscillating source that completes 116 cycles per minute. What is the

slava [35]

Answer:

$\lambda =6.32\ cm$

Explanation:

given,

number of cycle complete (f) = 116 cycles per minute

wavelength observed at 11 m in 1.5 m.

$v = \dfrac{distance}{time}$

$v = \dfrac{11}{1.5}$

v = 7.33 m/s

$\lambda = \dfrac{v}{f}$

$\lambda = \dfrac{7.33}{116}$

$\lambda =0.0632\ m$

$\lambda =6.32\ cm$

The wavelength of the wave is equal to $\lambda =6.32\ cm$

6 0

3 years ago

Which of the following would NOT produce a current? A. Moving a bar magnet inside a coil of wire O B. Moving a bar magnet toward

sineoko [7]

Answer:

Explanation:

7 0

3 years ago

How does the transfer of thermal energy occur?

natali 33 [55]

Heat<span> may be </span>transferred<span> by means of conduction, convection, or radiation. </span>

8 0

3 years ago

My buddy and I have just finished a dive to 15 metres/50 feet for 60 minutes. We want to return to the same site and depth and s

marishachu [46]

Answer:

1) Periodically check the no stop or NDL time on their computers

2) The dive computer planning mode can be used if available

3) Make use of a dive planning app

4) Check data from the RDP table or an eRDPML

Explanation:

The no stop times information from the computer gives the no-decompression limit (NDL) time allowable which is the time duration a diver theoretically is able to stay at a given depth without a need for a decompression stop

The dive computer plan mode or a downloadable dive planning app are presently the easiest methods of dive planning

The PADI RDP are dive planners based on several years of experience which provide reliable safety limits of depth and time.

7 0

2 years ago

How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800

ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

$A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2$

The inductance of the solenoid is given by;

$L = \frac{\mu_0 N^2A}{l} \\\\L = \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H$

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0

3 years ago