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UNO [17]
2 years ago
5

If you set up an experiment with two different independent variables, then the results would be_____.

Physics
1 answer:
aalyn [17]2 years ago
5 0

Answer:inconclusive

Explanation:

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Answer:

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4 0
2 years ago
A roller coaster track is 3000 meters long. It takes 100 seconds to travel once around the roller coaster. What is the average s
Svetllana [295]
All you have to do is divide 3000 by 100, Its 30
6 0
3 years ago
Read 2 more answers
Raina is running for student body president and doesn't want to forget her campaign speech. She practices her speech over and ov
mezya [45]

The answer is intentional.

6 0
3 years ago
Read 2 more answers
F = 2.0*10^20 N,
natka813 [3]

F=\dfrac{Gm_1m_2}{r^2}

With the given values of F,G,m_1,r, we have

2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}

Try dealing with the powers of 10 first: On the right, we have

\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}

Meanwhile, the other values on the right reduce to

\dfrac{6.67\times5.98}{3.8^2}\approx2.76

Then taking units into account, we end up with the equation

2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2

Now we solve for m_2:

m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}

m_2=7.25\times10^{22}\,\mathrm{kg}

or, if taking significant digits into account,

m_2=7.3\times10^{22}\,\mathrm{kg}

5 0
3 years ago
At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi
Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.  Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and \mu be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

                v = 0,     s = 84 m,     \mu = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

                      F = ma

                 -\mu N = ma

                 -\mu mg = ma

                      a = -\mu g

Also,    

               v^{2} = u^{2} + 2as

              (0)^{2} = u^{2} + 2(-\mu g)s

                  u^{2} = 2(\mu g)s

                            = \sqrt{2(0.36)(9.81 m/s^{2})(84 m)}

                            = 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

4 0
3 years ago
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