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vazorg [7]
1 year ago
13

A duck has a mass of 2.70 kg. As the duck paddles, a force of 0.110 N acts on it in a direction due east. In addition, the curre

nt of the water exerts a force of 0.220 N in a direction of 47.0° south of east. When these forces begin to act, the velocity of the duck is 0.120 m/s in a direction due east. Find (a) the magnitude and (b) the direction (relative to due east) of the displacement that the duck undergoes in 2.70 s while the forces are acting. (Note that the angle will be negative in the south of east direction.)
Physics
1 answer:
Maurinko [17]1 year ago
7 0

a) The magnitude will be 0.838m

b) The displacement will be -17.35°

<h3>What is displacement?</h3>

The path covered by an object from its initial point to final point.

Forces acting on the duck

x-axis:    0.13 + 0.16*cos(-56°) = 2.7 * ax  

ax = 0.0813 m/s^2

y-axis:    0.13*sin(-56°) = 2.7 * ay  

ay = -0.0491 m/s^2

The displacement on the x-axis

X = Vox * t + ax/2 * t²

X = 0.12* 3.2 + 0.0813/2*3.2²

X = 0.8

The displacement on the y-axis:

Y = Voy * t + ay/2 * t²

X = 0 - 0.0491/2*3.2²

Y=-0.25m

So, the magnitude and angle of this displacement [0.8,-0.25] is:

0.838m  at an angle of -17.35°

For more displacement related question visit

brainly.com/question/11934397

#SPJ1

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Answer:

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We know all the numbers on the right side,
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3 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
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