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3 years ago

Does heating a cup of Water allow it to dissolve more sugar ? Constant

Physics

2 answers:

RSB [31]3 years ago

7 0

Answer:

yes

Explanation:

Because the heat allows the the sugar to dissolve easily and stirring it causes it to dissolve even more

Anestetic [448]3 years ago

5 0

Answer:

Explanation:

Yes

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A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the

Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)

v₀ is the initial voltage on the cap

v is the voltage after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

τ = 20µ x 1.5k = 30 ms

v = v₀e^(t/τ)

0.5 = 1.5e^(t/30ms)

e^(t/30ms) = 10/3

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0

3 years ago

A sound has a sound level of 30 dB. Its intensity is what multiple of the standard reference level for intensities?

arsen [322]

<h2>Answer:</h2>

1000th multiple of the standard reference level for intensities.

<h2>Explanation:</h2>

The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;

β = 10 log (I / I₀) --------------------(i)

Where;

I₀ = reference intensity

Given from the question;

β = sound level = 30dB

Substitute this value into equation (i) as follows;

30 = 10 log (I / I₀)

Divide both sides by 3;

3 = log (I / I₀)

Take antilog of both sides;

10^(3) = (I / I₀)

1000 = I / I₀

Solve for I;

I = 1000I₀

Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)

7 0

3 years ago

Scientists have proven that genes play no role in self-esteem. Please select the best answer from the choices provided. T F

IRINA_888 [86]

The answer is False give thanks for the answer m8 and happy Halloween

6 0

3 years ago

Read 2 more answers

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago

Planet X is in a stable circular orbit around a star, as shown in the figure. Which of the following graphs best predicts the an

damaskus [11]

Answer:

Explanation:

Angular momentum is the product of moment of inertia and angular velocity.

L = I × ω

Since the planet follows a stable circular orbit, I and ω are constant and non-zero. Therefore, the angular momentum is constant and non-zero.

5 0

3 years ago