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3 years ago

What is the dependent variable? A. Speed B. Time C. Acceleration D. Distance

Physics

2 answers:

Zina [86]3 years ago

8 0

Time - i’ve done this

alexira [117]3 years ago

5 0

Answer: B. Time

Explanation: The speed in the independent variable representing the time taken to move somewhere. The time is the dependent variable because it changes on how fast or slow the vehicle moved. Hope this helped.

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How many significant digits are there in 11.090 mL? a. 5 b. 3 C. 1 d. 2

kati45 [8]

Answer:

Five

Explanation:

5 0

3 years ago

In places such as hospital operating rooms or factories for electronic circuit boards, electric sparks must be avoided. A person

FrozenT [24]

Time taken by the rubber-soled shoes to reduce a person's potential from 3.00×10³ V to 100V is 3.91s

The initial potential difference $\Delta V_o$ is related to the final potential difference $\Delta V$ by

$\Delta V=\Delta V_{o} e^{-t / \tau}$

Our target is to find t, so we rearrange equation (1) for t to be

$\Delta V=\Delta V_{o} e^{-t / RC}\\ \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) = \frac{-t}{RC}\\ t= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \tag{2}$

The body has a capacitance 150 pF while the foot has a capacitance 80 pF and both are in parallel connection. So, the equivalent capacitance is

$C= 150 \mathrm{~pF} + 80 \mathrm{~pF} = 230 \mathrm{~pF}$

The shoes of the rubber-soled has resistance $R=5000 \mathrm{~M\Omega}$ . Plug the values for $\Delta V, \Delta V_o$ and C into equation (2) to get the time t.

$t&= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \\ &= -( 5000 \times 10^{6} \mathrm{~\Omega})(230 \times 10^{-12} \mathrm{~F}) \ln \left ( \frac{100\mathrm{~V}}{3000 \mathrm{~V}} \right ) \\ &= {3.91 \mathrm{~s}}$

Learn more about capacitance here:

brainly.com/question/12644355

#SPJ4

3 0

2 years ago

You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen

Paladinen [302]

Answer:

(A)

$\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s \right )D+t_t^2v_s^2=0$

(B) D=54.71 m

Explanation:

Free Fall

When a particle is dropped in free air, it starts falling to the ground with an acceleration equal to the gravity. If one wanted to know the height of launching, it can indirectly be measured by the time it takes to reach the ground by the formula

$\displaystyle D=\frac{gt^2}{2}$

Solving for t

$\displaystyle t=\sqrt{\frac{2D}{g}}$

If we are taking into consideration the time we can hear the sound it makes when hitting the ground (or water in this case), we must also consider the speed of the sound for the time it takes to reach back our ears. That time can be computed from the basic equation for the speed

$\displaystyle t=\frac{D}{v_s}$

(A)

The total measured time is the sum of both times and it's given as $t_t=3.5\ seconds$

$\displaystyle t_t=\sqrt{\frac{2D}{g}}+\frac{D}{v_s}$

From this equation we'll manage to compute D

First, we isolate the square root

$\displaystyle \sqrt{\frac{2D}{g}}=t_t-\frac{D}{v_s}$

Let's square both sides

$\displaystyle \frac{2D}{g}=t_t^2-2t_t\frac{D}{v_s}+\frac{D^2}{v_s^2}$

Multiplying by $v_s^2$

$\displaystyle \frac{2Dv_s^2}{g}=t_t^2v_s^2-2t_tDv_s+D^2$

Rearranging and factoring

$\boxed{\displaystyle D^2-\left (\frac{2v_s^2}{g}+2t_tv_s\right )D+t_t^2v_s^2=0}$

Now, let's put in numbers:

$g=9.8\ m/s^2,\ v_s=345\ m/s,t_t=3.5\ sec$

$\displaystyle D^2-\left (\frac{2(345)^2}{9.8}+2(3.5)(345)\right )D+(12.25)345^2=0$

Computing all the coefficients:

$\displaystyle D^2-26,705.82D+1,458,056.25=0$

Solving for D, we have two possible solutions:

$D=54.71,\ D=26,651.11$

The second solution is called "extraneous", since it comes from squaring an equation, which can introduce non-valid (or external) solutions. It's impossible, given the conditions of the problem, that the well could be 26.5 km deep. So we'll keep the only solution as.

D=54.71 m

Let's prove our calculations by computing both times:

$\displaystyle t_1=\sqrt{\frac{2(54.71)}{9.8}}=3.34\ sec$

$\displaystyle t_2=\frac{54.71}{345}=0.16\ sec$

We can see their sum is 3.5 seconds, 3.34 of which were taken to reach the bottom of the well, and 0.16 sec took the sound to reach the top.

3 0

3 years ago

Exercise combined with a blank can help control weight

SIZIF [17.4K]

Answer:

Exercise combined with a diet can help manage weight, or you can contact a doctor

Explanation:

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4 years ago

While at a party, you pull up a sound intensity level app on your phone (everyone does stuff like that, right?), and it reads 83

allochka39001 [22]

To solve this problem it is necessary to apply the concepts related to Sound Intensity. The unit most used in the logarithmic scale is the decibel and mathematically this is expressed as

$\beta_{dB} = 10log_{10}\frac{I}{I_0}$