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pochemuha
3 years ago
14

Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s. (In order to b

etter see the limiting process keep at least three decimal places in your answer.)
Physics
1 answer:
weeeeeb [17]3 years ago
5 0

Answer:

4.408 m/s, 4.102 m/s, 4.026 m/s

Explanation:

The question is incomplete. The text of the original question states:

A race car moves such that its position fits the relationship

:

x=(4.0 m/s)t + (0.85 m/s^3) t^3

where x is measured in meters and t in seconds. Determine the instantaneous velocity of the car at t = 4.7 s, using time intervals of 0.40 s, 0.20 s, and 0.10 s.

We can find the instantanoues velocity of the car at any time t by calculating the derivative of the position, so we find:

v(t) = x'(t) = 4.0 m/s + 3\cdot (0.85 m/s^2) t^2 = 4.0 m/s + (2.55 m/s^2) t^2

And now we just need to substitute t=0.40 s, 0.20 s, and 0.10 s to find the corresponding instantaneous velocity:

v(0.40) = 4.0 + 2.55 (0.40)^2 = 4.408 m/s\\v(0.20) = 4.0 + 2.55 (0.20)^2 = 4.102 m/s\\v(0.10) = 4.0 + 2.55 (0.10)^2 = 4.026 m/s

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Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1320 K to a cold reservoir at 600 K. Calculate the
Zinaida [17]

Answer:0.0909 kJ/K

Explanation:

Given

Temperature of hot Reservoir T_h=1320 K

Temperature of cold Reservoir T_l=600 K

Heat of 100 kJ is transferred form hot reservoir to cold reservoir

Hot Reservoir is Rejecting heat therefore Q_1=-100 kJ

Heat is added to Reservoir therefore Q_2=100 kJ

Entropy change for system

\Delta s=\frac{Q_1}{T_1}+\frac{Q_2}{T_2}

\Delta s=\frac{-100}{1320}+\frac{100}{600}

\Delta s=-0.0757+0.1666=0.0909 kJ/K

As entropy change is Positive therefore entropy Principle is satisfied

         

4 0
3 years ago
What is the effect in the picture
stepan [7]
The effect of this problem is that negative particles and positive particles contract to each other caused by electrical force.
8 0
3 years ago
Lori wants to send a box of oranges to a friend by mail. The box of oranges cannot exceed a mass of 10.222 Kg. If each orange ha
Sergeu [11.5K]

Explanation:

Given that,

The box of oranges cannot exceed a mass of 10.222 Kg if we are sending to a friend by mail.

The mass of each orange is 198 g

We know that,

1 kg = 1000 g

10.222 kg = 10.222×1000 g

Let there are n number of oranges. So,

n=\dfrac{10.222\times 1000\ g}{198\ g}\\\\n=51.92\approx 52\ \text{oranges}

It means she can send 52 oranges and it is maximum quantity.

4 0
3 years ago
The distance from Abdullah's house to his school is 2.4km. Abdulla takes 0.6h to go to school on his cycle but takes only 0.4h t
vladimir1956 [14]

Answer:

The average speed can be calculated as the quotient between the distance travelled and the time needed to travel that distance.

To go to the school, he travels 2.4 km in 0.6 hours, then here the average speed is:

s = (2.4km)/(0.6 hours) = 4 km/h

To return to his home, he travels 2.4km again, this time in only 0.4 hours, then here the average speed is:

s' = (2.4 km)/(0.4 hours) = 6 km/h.

Now, if we want the total average speed (of going and returning) we have that the total distance traveled is two times the distance between his home and school, and the total time is 0.6 hours plus 0.4 hours, then the average speed is:

S = (2*2.4 km)/(0.6 hours + 0.4 hours)

S = (4.8km)/(1 h) = 4.8 km/h

5 0
2 years ago
Which one of the following is not equivalent to 2.50 miles?
lisabon 2012 [21]

Answer:

Choice C is not equivalent to 2.50 miles.

Explanation:

The given data is now converted into feet, inches, kilometers, yards and centimeters:

mi - ft

x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)

x = 13200\,ft

x = 1.320\times 10^{4}\,ft (Choice A)

mi - in

x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)\cdot \left(12\,\frac{in}{ft} \right)

x = 158400\,in

x = 1.584\times 10^{5} (Choice B)

mi - km

x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)

x = 4.025\,km (Different from Choice C)

mi - yd

x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi}\right) \cdot \left(\frac{1}{3}\,\frac{yd}{ft}  \right)

x = 4400\,yd

x = 4.40\times 10^{3}\,yd (Choice D)

mi - cm

x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)\cdot \left(100000\,\frac{cm}{km}\right)

x = 402500\,cm

x = 4.025\times 10^{5}\,cm (Choice E)

Choice C is not equivalent to 2.50 miles.

6 0
4 years ago
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