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Reika [66]
3 years ago
11

In the attached position versus time graph what is the magnitude of average velocity for the entire 19 seconds of the motion. An

swer in m/s and the number only to the
nearest 100th (ie. The second position after the decimal point.). Remind yourself of the definition of velocity.

Physics
1 answer:
evablogger [386]3 years ago
8 0

According to the plot, the positions at time <em>t</em> = 0 s and <em>t</em> = 19 s are -1 m and -2 m, respectively. So the average velocity for the 19-s interval is

v_{\rm ave} = \dfrac{-2\,\mathrm m - (-1\,\mathrm m)}{19\,\mathrm s} = -\dfrac1{19}\dfrac{\rm m}{\rm s}\approx \boxed{-0.05\dfrac{\rm m}{\rm s}}

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uppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviation of 0.8 ou
Brrunno [24]

Answer:

x= 9.53 ounces

Explanation:

Given that

Mean ,μ= 9 ounces

Standard deviation ,σ=0.8 ounces

He wants to sell only those potatoes that are among the heaviest 25%.

P=25% = 0.25

When P= 0.25 then Z=0.674

Lest take x is the the minimum weight required to be brought to the farmer's market.

We know that

x = Z . σ + μ

x= 0.674 ₓ 0.8 + 9 ounces

x= 9.53 ounces

3 0
3 years ago
How much time will it take to perform 440 joule of work at a rare of 11 w?​
kifflom [539]

Answer:

40sec

Explanation:

Data

Work = 440 J

Power= 11watt

time = ?

Power = work done/time

===> time = work done/power

= 440/11

= 40sec

7 0
3 years ago
How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v
777dan777 [17]
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
W= qV_i - qV_f
where q is the charge of the proton, q=1 e = 1.6\cdot 10^{-19}C, with e being the elementary charge, and V_i = +125 V and V_f = -55 V are the initial and final voltage.

Substituting, we get (in electronvolts):
W=e(125 V-(-55 V))=180 eV
and in Joule:
W=(1.6 \cdot 10^{-19})(125 V-(-55V))=2.88 \cdot 10^{-17}J

5 0
3 years ago
Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint
djyliett [7]

Answer:

F_5 >F_4>F_1 >F_2>F_3

Where F_i represent the force for each of the 5 cases i -1,2,3,4,5 presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

F = G \frac{m_{Earth} m_{Spaceship}}{r^2}

Where G is a constant G = 6.674 x10^{-11} m^2/ (ks s^2)

m_{Earth} represent the mass for the earth

m_{spaceship} represent the mass for the spaceship

r represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

F_5 >F_4>F_1 >F_2>F_3

Where F_i represent the force for each of the 5 cases i -1,2,3,4,5 presented on the figure attached.

6 0
3 years ago
An ambulance driver traveling at 31.0 m/s (69.3 mph) honks his horn as he sees a motorist ahead on the highway traveling in the
IrinaVladis [17]

Answer:

fo = 378.52Hz

Explanation:

Using Doppler effect formula:

f'=\frac{C-Vb}{C-Va}*fo

where

f' = 392 Hz

C = 340m/s

Vb = 20m/s

Va = 31m/s

Replacing these values and solving for fo:

fo = 378.52Hz

4 0
3 years ago
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