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nika2105 [10]
3 years ago
7

When a candle burns which forms of energy does the chemical energy in the candle change to

Physics
1 answer:
Gre4nikov [31]3 years ago
4 0
Thermal energy, radiant energy
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HELP!!
tresset_1 [31]

Answer:

So do 2400 divided by 70. I got 34.285714 and the numbers behind the decimal are repeating. If you round it you get 34.3

3 0
2 years ago
two pulse waves a and b arrive at a point in a medium simustaneously. if the amplitude of wave a is 35 units the amplitude of wa
Shalnov [3]
It depends on the type of interference.
For constructive interference, add the amplitudes to get |35 + 41| = 76 units.
For destructive, subtract them |35 - 41| = 6 units
3 0
3 years ago
Elements from opposite sides of the periodic table tend to form ________.
Ivenika [448]
Ionic compounds is your answer. What happens is one atom donates electron(s) to the other atom, making one positive and the other negative. The opposite atoms attract, forming an ionic bond. 

Hope this helps! :)
4 0
3 years ago
A bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes.
zaharov [31]

Given that,

Frequency emitted by the bat, f = 47.6 kHz

The speed off sound in air, v = 413 m/s

We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{413}{47.6\times 10^3}\\\\\lambda=0.00867\ m

or

\lambda=8.67\ mm

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.

3 0
3 years ago
A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the
andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

F = mg

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

\sum \tau = 0

Regarding the forces we have,

3Mg-1mg=0

Re-arrange to find M,

M = \frac{m}{3}

M = \frac{50}{3}

M = 16.67Kg

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0
3 years ago
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