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natima [27]
3 years ago
15

The force of friction between an object and the surface upon which it is sliding is 14N and the coefficient of friction between

them is 0.72. What is the weight of the object? ​
Physics
1 answer:
MrRa [10]3 years ago
4 0

Answer:

Explanation:

Given parameters:

Acceleration = 0.0m/w

Initial velocity = 0m/s

Final velocity  = 00m/s

Unknown:

Distance traveled  = ?

Solution:

To solve this problem, we use the expression below:

    q0   = y   + 0w

W is the final velocity

Y is the initial velocity

C is the acceleration

P is the distance

          0²  = 0²   +  (0 x 0.0 x S)

          000  = 00.0S

            S = 0m

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3 0
4 years ago
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about _____ years.
Airida [17]

Answer:

An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Explanation:

Given;

orbital period of 3 years, P = 3 years

To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.

Kepler's third law;

P² = a³

where;

P is the orbital period

a is the orbital semi-major axis

(3)² = a³

9 = a³

a = a = \sqrt[3]{9} \\\\a = 2.08 \ years

Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

5 0
3 years ago
You leave your house at 7:50
igor_vitrenko [27]

Answer:

LOL what?

Explanation:

3 0
3 years ago
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The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane
umka21 [38]

Answer:

\Delta p_x=7.75\times 10^{-25}\ kg-m/s

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, \Delta x=68\ pm=68\times 10^{-12}\ m

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }

Putting all the values, we get :

\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s

So, the momentum component of this electrons is greater than 7.75\times 10^{-25}\ kg-m/s.

6 0
4 years ago
This theory lost its appeal when astronomers discovered quasars and cosmic background radiation.
JulijaS [17]
<span>the Steady State Theory is the answer to this I believe!</span>
3 0
3 years ago
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