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Harman [31]
3 years ago
13

How much force is needed to accelerate a 66 kg skier at 2 m/sec^2?

Physics
1 answer:
Ymorist [56]3 years ago
5 0
The equation of force is M(mass)•A(acceleration)=132N
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What does it mean to have an acceleration of -10 m/s^2 when the velocity is 80 m/s?
mamaluj [8]

Answer:

1,3

Explanation:

As the acceleration is -10m/s^2 , that means deceleration is occurring. That means, the object is slowing down.

v=u-at

or, 0=80-10t

or, t=8 seconds

So, the object will stop in 8 seconds.

So, the correct answers are 1 and 3.

Hope, this helps you.

3 0
3 years ago
What is the SI unit of electric charge
shusha [124]

Answer:

The SI unit for electric chargeis the C (which is the abbreviation of Coulomb}

Explanation:

The SI unit for electric chargeis the Coulomb. The letter used is the C.

1 C = 1 As

4 0
4 years ago
A star has an absolute magnitude of 4 and a surface temperature of 5,000 degrees C. According to the HR diagram, list the type o
Ierofanga [76]
On sources it says it would  just be the super giant star 
3 0
3 years ago
You could use an elevator or the stairs to lift a box to tenth floor.which has greater power?
erastova [34]

Stairs don't have any power at all.  All the energy used to climb them
has to come from your muscles.

An elevator gets its power from the electric motors that lift it.  All YOU
have to do is stand there and look around.

All of this is a big part of the reason why elevators have become so
popular, and why no buildings with more than a few floors were built
before elevators were invented.


7 0
3 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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