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2 years ago

9

Two vectors have magnitudes 3 and 4 . how are the directions of the two vectors related if: a/the sum has magnitude 7.0

1 answer:

Marina CMI [18]2 years ago

5 0

Same directions
Perpendicular
Opposite directions

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The molecules of a gas are in constant random motion. This means that they have energy in what type of energy store?

serious [3.7K]

Answer:

Heat causes the molecules to move faster, (heat energy is converted to kinetic energy ) which means that the volume of a gas increases more than the volume of a solid or liquid.

Explanation:

3 0

3 years ago

The ratio of yellow bottles to green bottles is 3:7. If I have 50 bottles how many are yellow

azamat

15 yellow bottles.

3 + 7 = 10
50 / 10 = 5
3 x 5 = 15

7 0

2 years ago

17. Saan daw nakasakay ang mga Austronesyan nang dumating sa bansa? A. Bangka B. balangay C galyon D. barko

erastova [34]

Answer:

D.barko po

Explanation:

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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

Two blocks of clay, one of mass 1.00 kg and one of mass 7.00 kg, undergo a completely inelastic collision. Before the collision

RideAnS [48]

Answer:8 m/s

Explanation:

Given

$m_1=1 kg$

$m_2=7 kg$

kinetic Energy of $m_1=32 J$

initially $m_2$ is at rest and let say $m_1$ is moving with velocity u

kinetic Energy of $m_1 is =\frac{m_1u^2}{2}=32$

$u^2=64$

$u=8 m/s$

In Completely inelastic collision both mass stick together and move with common velocity

Suppose v is the common velocity

$m_1u+0=(m_1+m_2)v$

$v=\frac{1\times 8}{1+7}=1 m/s$

therefore Final velocity with which both blocks moves is 1 m/s

7 0

3 years ago