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Gelneren [198K]
3 years ago
14

Answers are - A 235 NB 376 NC 271 ND 188 NE 470 N

Physics
1 answer:
amm18123 years ago
8 0

Answer:

C

Explanation:

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Ymorist [56]

Answer: If you meant 5 miles in 20 minutes than it’s 1 mile in 5 minutes

Explanation:

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3 years ago
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Which is NOT a property of magnetic field lines?
salantis [7]
The answer would be B. Magnetic fields are invisible they cannot be directly observed
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A 5.45-g combustible sample is burned in a calorimeter. the heat generated changes the temperature of 555 g of water from 20.5°c
Y_Kistochka [10]
Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of  water

Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
    = (555 g)*(4.18 J/(°C-g)*(19 °C)
    = 44,078.1 J
    = 44,100 J (approximately)

Answer:  44,100 J

3 0
2 years ago
There are four springs stretched by the same mass.
brilliants [131]

Spring C stretches 100 cm.

Explanation:

The spring constant is simply the stiffness of the spring. The higher the spring constant the more stiff the spring is.

Spring constant shows the force needed to stretch a spring from it's equilibrium position. If a material requires more force to cause it to stretch, it will have a high spring constant.

          According to hooke's law "the force needed to extended an elastic material is directly proportional to its extension"

          F = ke

   k is the spring constant

   e is the extension

We see that the spring that stretches by 100 is the less stiff compared to other springs. It has the smallest spring constant.

Learn more;

Force brainly.com/question/8882476

#learnwithBrainly

8 0
2 years ago
Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
3 years ago
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