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Debora [2.8K]
4 years ago
11

Confused on b part i and ii and also c

Physics
1 answer:
34kurt4 years ago
4 0
B)

To calculate the values of for acceleration, it is helpful to remember what acceleration actually is.

Acceleration is the change in speed over time, with units \frac{m}{s^2}. The y-axis of this graph gives speed, and the x-axis gives time, so we find the point where the car is gaining speed, and we find the rate at which it gained speed.

i) We do this by calculating the slope. We obtain the points where the car starts and stops accelerating, (0,0),(20,40), and find the average rate of change (slope) between these two points.

m= \dfrac{y_2-y_1}{x_2-x_1} =  \frac{40}{20} = 2 \ m/s^2

ii) We do the same for the deceleration. We choose (40,40),(50,0).

m= \dfrac{y_2-y_1}{x_2-x_1}= \dfrac{-40}{50-40}=-4 \ m/s^2

c) We can most easily do this by calculating the distance for each segment of the graph.

For the first 20 seconds, the car moves at an average speed of 20 m/s. 20 \ s  \ \frac{20 \ m}{s} = 400 \ m

For the middle 20 seconds, the car moves at 40 m/s. 20 \ s  \ \frac{40 \ m}{s} = 800 \ m

For the final 10 seconds, the car moves at an average speed of 20 m/s. 10 \ s  \ \frac{20 \ m}{s} = 200 \ m

400+800+200=1400 \ m

You can also do this by finding the area under the graph. You don't need Calculus, because this isn't a curve!
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Y_Kistochka [10]

Answer:

Antarctic Circle

Explanation:

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Tropic of Capricorn Is it Southern Hemisphere counterpart, marking the most southerly position at which the Sun can be directly overhead.

7 0
3 years ago
Which calculates the intensity of an electric field at a point where a 0.50 C charge experiences a force of 20. N?
tester [92]

Answer: 40\ N/C

Explanation:

Given

Magnitude of charge is q=0.5\ C

Force experienced is F=20\ N

Electric field intensity is the electrostatic force per unit charge

\therefore E=\dfrac{F}{q}\\\\\Rightarrow E=\dfrac{20}{0.50}\\\\\Rightarrow E=40\ N/C

Thus, the electric field intensity is 40\ N/C

6 0
3 years ago
What temperature will 1L of H20 at 200°F become when a piece of copper,0.25kg at 260.928K, comes into contact with water?
Lady_Fox [76]

Answer:

Explanation:

mass of 1 L water = 1 kg .

200⁰F = (200 - 32) x 5 / 9 = 93.33⁰C .

260.928 K = 260.928 - 273 = - 12.072⁰C .

water is at higher temperature .

Let the equilibrium temperature be t .

Heat lost by water = mass x specific heat x  fall of temperature

= 1 x 4.2 x 10³ x ( 93.33 - t )

Heat gained by copper

= .25 x .385 x 10³ x ( t +  12.072 )

Heat lost = heat gained

1 x 4.2 x 10³ x ( 93.33 - t ) = .25 x .385 x 10³ x ( t +  12.072 )

93.33 - t = .0229 ( t + 12.072)

93.33 - t = .0229 t + .276

93.054 = 1.0229 t

t = 90.97⁰C .

7 0
3 years ago
A train travels 77 kilometers in 4 hours, and then 76 kilometers in 2 hours. What is its average speed?
Masja [62]

( (77/4) + 76/2 )/2 = 28.625 km/h is what i got

4 0
4 years ago
Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s
Vitek1552 [10]
The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for <span>52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=</span><span>binding energy
</span><span>52He-----------> 2 x 11p + 3 x10n is the equation bilan
</span>so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
                                                        mn=mass of neutron=<span>1.67 10^-27 kg
                                                        </span><span>m nucleus= 5
Dm= 2x</span>1.67 10^-27 kg+ 3x<span>1.67 10^-27 kg-5=  - 4.9 J
Eb= </span> - <span>4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so </span><span>-9.10^16 J/ 1.6 10^-19=  -5.625 10^35 eV

the final answer is </span><span>Eb /nucleon </span><span>=  -5.625 x10^35 eV</span>
8 0
3 years ago
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