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3 years ago

Lindsay and coworkers "slime in the first-grade teacher's desk" experiment showed that presenting

1 answer:

4 0

The answer that completes the statement is this: A PHOTOGRAPH OF THE PARTICIPANT'S FIRST-GRADE CLASS INCREASED THE LIKELIHOOD OF FALSE MEMORIES. This experiment is similar to Maryanne Gary's experiment which also states that viewing memory as a tape or video recording can be mistaken and may create false memories.

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Can i zoom with somebody

stealth61 [152]

Answer:

If not because you know you just can't

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2 years ago

A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational

Pavlova-9 [17]

<span> gravitational force varies based on 1/r^2
when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.
</span><span>As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)
</span>hope this helps

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3 years ago

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A box is being pulled to the right. What is the magnitude of the Kinect frictional force?

Anna35 [415]

The answer to this question is A - 25 N

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3 years ago

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What causes dispersion of light?

AysviL [449]

Dispersion occurs due to the different degrees of refraction experienced by different colours of light. Light of different colours may travel with the same speed in a vacuum, but they travel at different speeds in some refracting medium. The speed of violet light is relatively lower than that of red light.

7 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago