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olga55 [171]
3 years ago
13

Which is an example of radiation?

Physics
2 answers:
oksano4ka [1.4K]3 years ago
5 0
I believe The snowman should be the answer
marishachu [46]3 years ago
5 0
The correct answer is the snowman!
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A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa
Lesechka [4]
At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
U_i= \frac{1}{2} ka^2
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
K_i =0
And the total energy of the system is
E_i = U_i+K= \frac{1}{2}ka^2

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
U_f = 0
while the mass is moving at speed v, and therefore the kinetic energy is
K_f =  \frac{1}{2} mv^2
And the total energy is
E_f = U_f + K_f =  \frac{1}{2} mv^2

For the law of conservation of energy, the total energy must be conserved, therefore E_i = E_f. So we  can write
\frac{1}{2} ka^2 =  \frac{1}{2}mv^2
that we can solve to find an expression for v:
v= \sqrt{ \frac{ka^2}{m} }
6 0
3 years ago
What do you want to learn about simple machines? (could someone come up with a question about simple machines for me)​
bezimeni [28]

Answer:

what are simple machines lol

Explanation:

6 0
3 years ago
Use Percentages Given that the molecular mass of magnesium
ira [324]

Answer:

27%

Explanation:

15.999 divided by 58.32 = .27433128

Move the decimal place over 2 places.

27%

5 0
4 years ago
The distance-time graph for a faster moving object has a smaller slope than the graph for a slower moving object. True or false
allsm [11]
False because faster moving slope would be going up and slower would be going down because its decreasing 
6 0
3 years ago
Read 2 more answers
Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which th
chubhunter [2.5K]

Answer:

<em>1.01 W/m</em>

Explanation:

diameter of the pipe d = 30 mm = 0.03 m

radius of the pipe r = d/2 = 0.015 m

external air temperature Ta = 20 °C

temperature of pipe wall Tw = 150 °C

convection coefficient at outer tube surface h = 11 W/m^2-K

From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.

area of the pipe per unit length A = \pi r ^{2} = 7.069*10^{-4} m^2/m

convectional heat loss Q = Ah(Tw - Ta)

Q = 7.069 x 10^-4 x 11 x (150 - 20)

Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>

4 0
4 years ago
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