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2 years ago

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Making the material you are trying to memorize personally meaningful to you is called the ____________.

1 answer:

____ [38]2 years ago

8 0

Answer:

Making the material you are trying to memorize personally meaningful to you is called the?

c. self-referencing effect.

xXxAnimexXx

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Calculate the value of 200°C in Kelvin

Elan Coil [88]

Answer:

473.15

Explanation:

3 0

2 years ago

A 16-cm-long straight line connects the center of a turntable to its edge. The turntable rotates counter-clockwise at 45 rpm. A

Bond [772]

Answer:

$\mathbf{V_x = 3.25 \ cm/s}$

$\mathbf{V_y = 1.29\ cm/s}$

Explanation:

Given that:

The radius of the table r = 16 cm = 0.16 m

The angular velocity = 45 rpm

= $45 \times \dfrac{1}{60}(2 \pi)$

= 4.71 rad/s

However, the relative velocity of the bug with turntable is:

v = 3.5 cm/s = 0.035 m/s

Thus, the time taken to reach the bug to the end is:

$t = \dfrac{r}{v}$

$t = \dfrac{0.16}{0.035}$

t = 4.571s

So the angle made by the radius r with the horizontal during the time the bug gets to the end is:

$\theta = \omega t$

$\theta = 4.712 \times 4.571$

$\theta = 21.54^0$

Now, the velocity components of the bug with respect to the table is:

$V_x = Vcos \theta$

$V_x = 0.035 \times cos (21.54^0)$

$V_x = 0.0325 \ m/s$

$\text {V_x = 3.25 \ cm/s}$ $\mathbf{V_x = 3.25 \ cm/s}$

Also, for the vertical component of the velocity $V_y$

$V_y = V sin \theta$

$V_y = 0.035 \times sin (21.54^0)$

$V_y = 0.0129\ m/s$

$\mathbf{V_y = 1.29\ cm/s}$

6 0

2 years ago

Abishek is a runner. He runs the 100 m sprint

konstantin123 [22]

Answer:

1.67m/s

Explanation:

Total Distance to be travelled by a Runner=100m

Time Taken=10*6s

Speed=Distance/Time

=100/10*6=10/6=1.67m/s

3 0

3 years ago

Read 2 more answers

An LC circuit consists of a 3.4-µF capacitor and a coil with a self-inductance 0.080 H and no appreciable resistance. At t = 0 t

alexira [117]

Answer

given,

capacitance = C = 3.4-µF

inductance = L = 0.08 H

frequency is expressed as

$f = \dfrac{1}{2\pi\sqrt{LC}}$

time period

$T = \dfrac{1}{f}=2\pi\sqrt{LC}$

after time T/4 current reach maximum

$t = \dfrac{T}{4}$

$t = \dfrac{2\pi\sqrt{LC}}{4}$

$t = \dfrac{2\pi\sqrt{0.08 \times 3.4 \times 10^{-6}}}{4}$

t = 8.2 x 10⁻⁴ s

t = 0.82 ms

b) using law of conservation

$\dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2$

$I^2 = \dfrac{CV^2}{L}$

$I^2 = \dfrac{C}{L}\dfrac{Q^2}{C^2}$

$I =\sqrt{\dfrac{Q^2}{CL}}$

$I =\sqrt{\dfrac{(5.4 \times 10^{-6})^2}{0.08 \times 3.4 \times 10^{-6}}}$

I = 0.010 A

I = 10 mA

4 0

3 years ago

What will happen to the current if the voltage is reduced to one half?

stira [4]

We use v=IR and assuming the resistance doesn’t change we can also say that the voltage and current (I) are directly proportional which means the voltage also decreases by 1/2

8 0

2 years ago