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Leto [7]
3 years ago
8

A controlled scientific investigation determines the effect of a(n) _______ variable in an experiment.

Physics
1 answer:
ollegr [7]3 years ago
8 0
I believe it is dependent :)
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Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute and com
____ [38]

Answer:

Solid State or Condense matter book

Explanation:

8 0
4 years ago
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What type(s) of energy is/are contained in a roller coaster car after it has come to a complete stop at the station? A. kinetic
blondinia [14]

Answer:

I think it is just potential

Explanation:

If it’s right pls mark the brainliest

7 0
3 years ago
Based on the simple blackbody radiation model described in class, answer the following question. The planets Mars and Venus have
Lera25 [3.4K]

Answer:

The extent of greenhouse effect on mars is G_m =  87 K  

Explanation:

From the question we are told that

     The albedo value of Mars is  A_1  = 0.15

      The albedo value of Mars is  A_2  = 0.15

       The surface temperature of Mars is  T_1 = 220 K

        The surface temperature of Venus is  T_2 = 700 K

          The distance of Mars from the sun is d_m = 2.28*10^8 \ km = 2.28*10^8* 1000 = 2.28*10^{11} \  m

          The distance of Venus from sun is  d_v = 1.08 *10^{8} \ km = 1.08 *10^{8} * 1000 =  1.08 *10^{11} \ m

       The radius of the sun is R = 7*10^{8} \ m

        The energy flux is   E = 6.28 * 10^{7} W/m^2

The solar constant for Mars is mathematically represented as

 

          T = [\frac{E R^2 (1- A_1)}{\sigma d_m} ]

Where \sigma is the Stefan's constant with a value  \sigma = 5.6*10^{-8} \ Wm^{-2} K^{-4}

So substituting values

         T = \frac{6.28 *10^{7} * (7*10^8)^2 * (1-0.15)}{(5.67 *10^{-8}) * (2.28 *10^{11})^2)}

          T = 307K

So the greenhouse effect on Mars is  

           G_m =  T -  T_1

           G_m =  307 - 220

          G_m =  87 K

   

3 0
4 years ago
On a trip to the Colorado Rockies, you notice that when the freeway goes steeply down a hill, there are emergency exits every fe
Zanzabum

Answer:

The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.

Explanation:

4 0
3 years ago
A scientist fixes a test charge q′ to point P and then measures the electrostatic force it experiences there. She then calculate
nordsb [41]

Answer:

<em>Correct answer: B.</em>

Explanation:

Electrostatic Field

It measures the electric effect of a charge distribution in its surroundings. If we wanted to test or measure the electric field by using a point-charge of value q', then the electric field is

\displaystyle E=\frac{F}{q'}

The electrostatic force between two point charges is

\displaystyle F=\frac{k\ q_1\ q'}{r^2}

So, the electric field is

\displaystyle E=\frac{\frac{k\ q_1\ q'}{r^2}}{q'}

\displaystyle E=\frac{k\ q_1}{r^2}

The theory shows that the electric field doesn't depend on the test charge used, i.e., if we now use q'=-q', the electric force will be of the same magnitude but in the opposite direction, thus the electric field will be the same. Let's recall the formula is used to compute the scalar value of the field, the direction must be studied separately.

Now, if we changed the test charge to another value, say 2q, the measured force will be

\displaystyle F=\frac{k\ q_1\ 2q}{r^2}

And the new electric field is

\displaystyle E=\frac{k\ q_1}{r^2}

We can see the electric field is not affected by the value of the test charge.

Finally, if we move the test charge to another location and keep the same charge, the electric force will vary and the electric field will be different.

Correct answer: B.

7 0
3 years ago
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