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uysha [10]
3 years ago
6

Which needs less heat to increase its temperature,

Physics
2 answers:
rosijanka [135]3 years ago
6 0
Sand is what it said and I think I agree

Julli [10]3 years ago
5 0

Answer:

sand

Explanation:

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Which best describes a difference between laser light and regular light?
Mashcka [7]

Answer:

B

Explanation:

6 0
2 years ago
Why are the eggs and confectioner’s sugar amounts easy to work with to make 8 brownies?
Orlov [11]

Answer:

Because they are both easy to measure (?)

Explanation:

(I'm not really sure, there are no choices. If there were different options I might be able to better answer this)

6 0
3 years ago
HELP...<br> 3.00 amu = _____ Mev.<br><br> 3.22 x 10-3<br> 2.79 x 103<br> 3.10 x 102
DanielleElmas [232]
The correct answer is:
2.79 \cdot 10^3 MeV
Let's see why.

1 amu corresponds to the mass of the proton, which is:
m_p = 1.66 \cdot 10^{-27} kg
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J
Now we can convert it into electronvolts:
E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
3 amu = 3 \cdot 934 MeV  \sim 2790 MeV = 2.79 \cdot 10^3 MeV
5 0
3 years ago
Read 2 more answers
A car has a momentum of 20,000 kg. m/s. What would the car's momentum be if its velocity doubles?
Savatey [412]

Answer:

40,000

Explanation:

Momentum is defined as mass*velocity, so a doubling of velocity means a doubling of momentum

8 0
3 years ago
ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit
horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

F = \dfrac{d^4G\delta}{8D^3N}

\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm

F = \dfrac{d^4G}{8D^3}\times 0.004

F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004

     F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

\tau = \dfrac{8FDk_s}{\pi d^3}

\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}

              = 442.6 Mpa

The tensile strength of the steel material of  ASTM A229 is equal to 1300 Mpa

now,

\tau_s \leq 0.45 S_u

\tau_s \leq 0.45 \times 1300

\tau_s \leq 585\ Mpa

since corrected stress is less than the \tau_s

hence, spring will return to its original shape.

6 0
3 years ago
Read 2 more answers
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