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3 years ago

If C moves to position 5 (60 meters), what is its average velocity during these 5 seconds?

2 answers:

4 0

The average velocity is simply the change in displacement over the specified time interval. In this case we have a total of five intervals, which represents 5 seconds. Dividing 60 m by 5s, gives us 12 m/s, which is the magnitude of the average speed in the interval.

Hoochie [10]3 years ago

3 0

Velocity:

60 meters/ 5 seconds= 12 meters/second

Velocity=12meters/second

(60 meters divided by 5 seconds equals 12 meters per second)

Hope this helps!

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A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

3 years ago

Question 4. A tuning fork ‘A’ produces 6 beats/sec with another fork ‘B’ of un-known frequency. On

8090 [49]

Clever problem.

We know that the beat frequency is the DIFFERENCE between the frequencies of the two tuning forks. So if Fork-A is 256 Hz and the beat is 6 Hz, then Fork-B has to be EITHER 250 Hz OR 262 Hz. But which one is it ?

Well, loading Fork-B with wax increases its mass and makes it vibrate SLOWER, and when that happens, the beat drops to 5 Hz. That means that when Fork-B slowed down, its frequency got CLOSER to the frequency of Fork-A ... their DIFFERENCE dropped from 6 Hz to 5 Hz.

If slowing down Fork-B pushed it CLOSER to the frequency of Fork-A, then its natural frequency must be ABOVE Fork-A.

The natural frequency of Fork-B, after it gets cleaned up and returns to its normal condition, is 262 Hz. While it was loaded with wax, it was 261 Hz.

4 0

3 years ago

The volumes of the block is 10,500 mL and density is 8.52 g/mL. Find the block's mass Don't forget units and a space

Schach [20]

Answer:

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 8.52 × 10,500

We have the final answer as

Hope this helps you

3 0

2 years ago

Consider two less-than-desirable options.

Gemiola [76]

Answer:

The force would be the same in both cases - option C.

Explanation:

The change in momentum is known as an impulse. In the two cases under consideration, the change in momentum is the same, thus impulse for both cases is the same.

Impulse is the average force multiplied by time interval.

I = F(average)*ΔT. Where F(average) is the average force and ΔT is the time interval.

The average force in both cases is the same since the collision time is the same.

Thus option C is the correct answer.

7 0

3 years ago

If you drop an object from a height of 1.9 m, it will hit the ground in 0.62 s. if you throw a baseball horizontally with an ini

faltersainse [42]

That's the cool thing about free fall. The amount of time it takes to fall remains the same.

In this case, a ball that is simply dropped from rest will fall at the same rate as a ball that had some umph in the horizontal direction.

6 0

2 years ago