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3 years ago

What investigations are best for demonstrating cause and effect relationship

Physics

2 answers:

poizon [28]3 years ago

8 0

The cause would be how the idea started or developed and the effect would be how they both effected you together

ella [17]3 years ago

7 0

Answer:

Experimental investigations are best for demonstrating cause-and-effect relationships.

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A 2kg object travels to the right at 5.5 m/s after colliding with a 3kg object, initially at rest, what are the speeds of each o

Snowcat [4.5K]

Answer:

Velocity of Object with 2 kg= 3.390 m/s

Velocity of Object with 3 kg= 3.404 m/s

Explanation:

From the picture, it can be seen that object B is initially at rest while object A is travelling at a speed of 5m/s. After the collision, Object A moves at an angle of 65 degrees while object B moves at an angle of 37 degrees.

We also know that momentum of a closed system is conserved.

Initial momentum along the x-axis = 2*5.5 = 11

Initial momentum along y-axis = 0

Final momentum along x-axis= a*Cos(65)*2 +b*Cos(37) *3= 11 (a is the velocity of object A of 2 kg after collision where as b is the velocity of object B of 3 kg after collision. velocity is multiplied by cosines of the angle from x axis to give the horizontal component of the velocities).

Final momentum along y-axis = a*Sin(65)*2 - b*Sin(37)*3 =0 (We can see that vertical components of velocity are opposite in direction to each other)

Solve both the equations simultaneously for a and b.

3 0

4 years ago

Colors seen on the cover of our physics book result from color Group of answer choices addition. subtraction. either of these ne

Alex Ar [27]

Colors seen on the cover of our physics book result from color is due to Subtraction.

What is physics of color subtraction?

Some visible spectrum wavelengths are intentionally removed during the subtraction procedure.
For instance, the yellow filter transmits the green and red colors while blocking the blue.
Red and blue are transmitted while the green is blocked by the magenta filter.
Blue and green are transmitted while red is blocked by the cyan filter.
Subtractive mixing gets its name from the fact that when colors are mixed, wavelengths are removed from what we see because each paint absorbs some of the wavelengths that the other paint reflects, leaving us with less wavelengths afterward.

Learn more about Subtractive mixing with the help of the given link:

brainly.com/question/1871483

#SPJ4

6 0

2 years ago

How far can a person run in 15 minutes if he or she runs at an average speed of 16 km/hr

Studentka2010 [4]

4 km/hr. There are four quarters in an hour, so just divide 16 by 4 and you get 4. It is directly proportional.

6 0

3 years ago

two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will h

Aneli [31]

Answer:

On that line segment between the two charges, at approximately $0.7\; \rm m$ away from the smaller charge (the one with a magnitude of $5 \times 10^{-19}\; \rm C$ ,) and approximately $1.3\; \rm m$ from the larger charge (the one with a magnitude of $20 \times 10^{-19}\; \rm C$ .)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let $k$ denote the Coulomb constant, and let $q$ denote the size of a point charge. At a distance of $r$ away from the charge, the electric field due to this point charge will be:

$\displaystyle E = \frac{k\, q}{r^2}$ .

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let $q_1 = 5\times 10^{-19}\; \rm C$ and $q_2 = 20 \times 10^{-19}\; \rm C$ . Assume that the electric field is zero at $r$ meters to the right of the $5\times 10^{-19}\; \rm C$ point charge. That would be $(2 - r)$ meters to the left of the $20 \times 10^{-19}\; \rm C$ point charge. (Since this point should be between the two point charges, $0 < r < 2$ .)

The electric field due to $q_1 = 5\times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}$ .

The electric field due to $q_2 = 20 \times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}$ .

Note that at all point in this section, the two electric fields $E_1$ and $E_2$ will be acting in opposite directions. At the point where the two electric fields balance each other precisely, $| E_1 | = | E_2 |$ . That's where the actual electric field is zero.

$| E_1 | = | E_2 |$ means that $\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}$ .

Simplify this expression and solve for $r$ :

$\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0$ .

$\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0$ .

Either $r = -2$ or $\displaystyle r = \frac{2}{3}\approx 0.67$ will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, $0 < r < 2$ . Therefore, $(-2)$ isn't a valid value for $r$ in this context.

As a result, the electric field is zero at the point approximately $0.67\; \rm m$ away the $5\times 10^{-19}\; \rm C$ charge, and approximately $2 - 0.67 \approx 1.3\; \rm m$ away from the $20 \times 10^{-19}\; \rm C$ charge.

8 0

3 years ago

A cameraman sitting near the open door of a news helicopter accidentally drops his 140-g mobile phone out

Agata [3.3K]

Answer:B.140 m/s

Explanation:

4 0

4 years ago

Read 2 more answers