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3 years ago

Finish the sentence below with the word that makes the sentence true.

Physics

1 answer:

Keith_Richards [23]3 years ago

7 0

Answer:

Animals must eat other plants or animals to obtain the <u>energy</u> stored in the food

Explanation:

One classification of living organisms, according to the way they obtain energy, is that of autotrophs and heterotrophs. The first group is represented by plants, which process their own nutrients from inorganic matter.

<u>Animals -heterotrophes- are unable to process their own nutrients</u>, so they must obtain them from other organisms, either plants or animals. External food sources provide them with nutrients, which contain the energy substrate needed to perform their vital functions.

Learn more:

Autotrophs and heterotrophs brainly.com/question/7695115

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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)

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

Explanation:

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.

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

(e) For photons of energy 7.10 eV, what stopping potential would be required to arrest the current of photoelectrons

Masja [62]

2.37eV stopping potential would be required to arrest the current of photoelectrons.

<h3 /><h3>What is stopping potential ?</h3>

The minimal negative voltage that must be provided to the anode to halt the photocurrent is known as stopping potential. When expressed in electron volts, the maximal kinetic energy of the electrons is equal to the stopping voltage.

Kmax = eV₀

2.37eV = eV₀

V₀ = 2.37eV

to learn more about stopping potential go to - brainly.com/question/4655588

#SPJ4

8 0

1 year ago

Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st

DedPeter [7]

Answer:

E1 = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q) = +11nC

distance between thin glass rods = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = $\frac{Q}{2\pi e_{0}rL }$

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.01} - \frac{1}{0.03} )$ ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

= $\frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )$

= 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.02} - \frac{1}{0.02} )$

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

4 0

2 years ago

A wave has angular frequency 30 rad/s and wavelength 2.0 m. What are its (a) wave number and (b) wave speed

masya89 [10]

Answer:

a) 15 b) 60 i think is the answer

4 0

1 year ago

Frame S' passes frame S in the usual way (positive directions). An object also moves in the positive direction. Which is true? T

lisov135 [29]

Answer:

The answer is "The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values."

Explanation:

The S' frame and the object are moving in a positive direction. The object is moving with respect to the S frame so the S frame the rest frame

take the velocity of the object with respect to the rest frame as v and the velocity of the S' frame with respect S frame as v2

relative velocity of the object to the S' frame would be

Vrel = v2- v

This means the Vrel of the object with respect to the S' frame is less than the Vrel of the object with respect to the S frame

However is the S' velocity is greater than that of the object then the Vrel of the object with respect to the S' frame is greater than the Vrel of the object with respect to the S frame.

This would mean the second option is the answer, the relative speed of the object depends on the actual values.

3 0

3 years ago