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2 years ago

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❤️❤️❤️❤️WILL MARK BRAINLIEST PLEASE ANSWER BOTH PARTS!❤️❤️❤️❤️

1 answer:

alekssr [168]2 years ago

8 0

Answer:

the answer to the first one is B and the second one is c I hope I got them right

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Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

nignag [31]

The position of the particle is given by:

x(t) = t³ - 12t² + 21t - 9

Differentiate x(t) with respect to t to find the velocity x'(t):

x'(t) = 3t² - 24t + 21

Differentiate x'(t) with respect to t to find the acceleration x''(t):

x''(t) = 6t - 24

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2 years ago

If a baseball has a negative velocity and a negative acceleration, its speed is

Korvikt [17]

Answer:constant cause it keeps happening Or it might be decreasing but I’m not sure

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2 years ago

The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of

kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

$\rho_{0\°C}= 730 kg/m^{3}$ is the density of gasoline at $0\°C$

$\beta=9.60(10)^{-4} \°C^{-1}$ is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to $m^{3}$ :

$V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3}$ (1)

Knowing density is given by: $\rho=\frac{m}{V}$ , we can find the mass $m_{1}$ of 8.50 gallons:

$m_{1}=\rho_{0\°C}V$

$m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg$ (2)

Now, we have to calculate the factor $f$ by which the volume of gasoline is increased with the temperature, which is given by:

$f=(1+\beta(T_{f}-T_{o}))$ (3)

Where $T_{o}=0\°C$ is the initial temperature and $T_{f}=21.7\°C$ is the final temperature.

$f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C))$ (4)

$f=1.020832$ (5)

With this, we can calculate the density of gasoline at $21.7\°C$ :

$\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)$

$\rho_{21.7\°C}=745.207 kg/m^{3}$ (6)

Now we can calculate the mass of gasoline at this temperature:

$m_{2}=\rho_{21.7\°C}V$ (7)

$m_{2}=(745.207 kg/m^{3})(0.0323 m^{3})$ (8)

$m_{2}=24.070 kg$ (9)

And finally calculate the mass difference $\Delta m$ :

$\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg$ (10)

$\Delta m=0.4911 kg$ (11) This is the extra mass of gasoline

6 0

3 years ago

What will happen to plant height if the amount of available light is reduced due to global dimming? Which type of investigation

Lorico [155]

The plant will not grow. In fact it could have all the nutrients and all the water it needs, but without a sufficient amount of light, it could die because its leaves are meant for a certain minimum amount of light.

I'll come back and see if you have posted the question you wanted and edit my answer.

6 0

3 years ago

how much force would be required to produce 88 j of work when pushing a box 1.1meters at an angle of 10 degrees?

ycow [4]

Answer:81.235N

Explanation:

Work=88J

theta=10°

distance=1.1 meters

work=force x cos(theta) x distance

88=force x cos10 x 1.1 cos10=0.9848

88=force x 0.9848 x 1.1

88=force x 1.08328

Divide both sides by 1.08328

88/1.08328=(force x 1.08328)/1.08328

81.235=force

Force=81.235

5 0

2 years ago