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aleksklad [387]
3 years ago
5

You are an astronaut floating alongside your space ship. If you throw a baseball with a speed of 10 m/s, you will move at:

Physics
1 answer:
ZanzabumX [31]3 years ago
4 0

Answer:

D

Explanation:

There is no friction to stop you from moving BC you are in space, however you have a larger mass than the ball, so it takes more force to get you up to the same speed as the baseball. You will move in the opposite direction of the ball because you exerted force on the ball in one direction and therefore yourself in the opposing direction.

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Will mark as brainliest if correct!!!!!!!!!!!!!!!!!!!!!!!!!
Digiron [165]

Answer:

Incident ray

Explanation:

"To describe the reflection of light, we will use the following terminology. The incoming light ray is called the incident ray. The light ray moving away from the surface is the reflected ray. The most important characteristic of these rays is their angles in relation to the reflecting surface."

https://www.siyavula.com/read/science/grade-11/geometrical-optics/05-geometrical-optics-03

3 0
3 years ago
Read 2 more answers
To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv
Lesechka [4]

Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

        E = k λ (-1/x){(-1/x)}^{d+L} _{d}

   

Evaluating

        E = k λ [ 1/d  - 1/ (d+L)]

Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

3 0
3 years ago
n an object. one force is 3N to the east and the other force is 9n to the west. what is the net force acting on the object ​
cricket20 [7]

Answer:

-6N

Explanation:

The force to the east is acting in the positive x-direction therefore it is positive. The force to the east is in the negative x-direction therefore it is negative. The net force is just the sum of the two so 3-9=-6

4 0
2 years ago
A series of optical telescopes produced an image that has a resolution of about 0.00350 arc second.
Mila [183]

Answer:

The smallest diameter is D =122 \ m

Explanation:

From the question we are told that

       The resolution of the telescope is \theta  =  0.00350 \ arc \ second

           The wavelength is  \lambda = 1.70 \mu m = 1.70 *10^{-6} \ m

From the question we are told that

        1 arc \ sec = \frac{1}{3600^o}

So      0.00350 \ arc \ second = x

Therefore

             x =  0.00350  *  \frac{1}{3600 }

              x = ( 9.722*10^{-7} )^o

Now  1^o  =  \frac{\pi}{180}

   So  (9.722*10^{-7})^o =  \theta

  =>    \theta  =  (9.722*10^{-7}) * \frac{\pi}{180}

           \theta  =  1.69*10^{-8} rad

The smallest diameter is mathematically represented  as

          D = \frac{1.22 \lambda }{\theta  }

substituting values

           D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8}  }

           D =122 \ m

   

6 0
3 years ago
Give 5 examples of physical properties that could be used to describe a solid
ycow [4]
1.) appearance
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