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3 years ago

Which is a unit of volume? A) centimeter B) meter C) milligram D) milliliter

Physics

1 answer:

german3 years ago

6 0

Answer:

I think milligram is the correct answer.

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A rocket burns fuel to create hot gases that explode violently out of the rocket engine. This explosion creates thrust. Thrust i

Lerok [7]

Answer:

Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Explanation:

From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

5 0

3 years ago

James weighs 67 pounds on Earth. He has a mass of 30 kilograms. Which of the following is true?

morpeh [17]

Answer:

I think its B. James would have a smaller mass on the Moon than he does on Earth.

sorry if i did it wrong

4 0

3 years ago

Read 2 more answers

The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.55 g of conden

SSSSS [86.1K]

Answer:

m = 62.14 g

Explanation:

Energy used to melt the ice is the energy released by the condensation of the water forms on the glass

so here we have

energy for the condensation of water is given as

let mass of water condensed = m

$E = m_1 L_f$

now the energy of vaporization is given as

$E = m_2 L_v$

here we know that

$L_f = 79.8 kCal/kg$

$L_v = 580 k Cal/kg$

Now we have

$8.55 \times 580 = m \times 79.8$

$m = 62.14 g$

3 0

3 years ago

calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2

igor_vitrenko [27]

First, balance the reaction:

_ KClO₃ ==> _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃ ==> 2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃ ==> 2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0

3 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

3 years ago