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shtirl [24]
3 years ago
8

4. If you swing wide to the left before turning right, another driver may try to pass you on the right. True or False

Physics
1 answer:
olchik [2.2K]3 years ago
8 0

Answer:

True.

Explanation:

Don't turn wide to the left as you start the turn. A driver behind may think you are turning left and try to pass you on the right. You may crash into the other vehicle as you complete your turn.

Instead, slowly give yourself and others more time to avoid problems, keep the rear of the vehicle close to the curb. This will stop other drivers from passing you on the right. This is called (button Hook)

If you are driving a truck or bus that cannot make the right turn without swinging into the other lane, turn wide as you complete the turn.

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Identify the word being referred to choose your answer from the words below​
SSSSS [86.1K]

Answer:

1:Rotation

2:Axis

3:Aphelion

4:orbit

7 0
3 years ago
If we could easily create machines or systems with no friction, what could we use them for?
Alexus [3.1K]

Answer:

We could make a frictionless generator.

Explanation:

This would change life by making the cost for energy next to zero

3 0
2 years ago
Sort the chemical equations based on the chemical reactions they represent.
Svetradugi [14.3K]

Decomposition reactions are said to be those reactions in which a reactants breakdown into two or more products. The general reaction for decomposition reactions is as follow,

                                         ABC    →    A  +  B  +  C

Specific Examples are as,

                                Water    →     Hydrogen + Oxygen

                                2 H₂O    →           2 H₂     +      O₂

               Calcium carbonate    →     Calcium oxide + Carbon dioxide

                            CaCO₃              →              CaO         +        CO₂

While, Synthetic reactions are said to be those reactions in which two or more reactants combine to form two or more products. The general reaction for synthetic reactions is as follow,

                                         A  +  B  +  C    →    ABC

Specific Examples are as,

                                  Iron + Oxygen     →     Iron Oxide

                                    2 Fe  +  3 O₂      →        2 Fe₂O₃

                          Sodium + Chlorine     →     Sodium chloride

                              2 Na  +       Cl₂        →            2 NaCl

                            Sulfur + Oxygen      →      Sulfur dioxide

                                  S    +    O₂            →             SO₂

                   Potassium + Chlorine      →      Potassium chloride

                            2 K     +     Cl₂            →              2 KCl

4 0
3 years ago
Read 2 more answers
A rocket sled with an initial mass of 3 metric tons, invluding 1 ton of fuel, rests on a level section of track. At t=0, the sol
ELEN [110]

Answer:

v = 719.2 m / s and     a = 83.33 m / s²

Explanation:

This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is

           v - v₀ = v_{e} ln (M₀ / M)

where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket

In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s

            m_fuel = 75 10

            m_fuel = 750 kg

As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is

              M = 3000 -750 = 2250 kg

let's calculate

            v - 0 = 2500 ln (3000/2250)

            v = 719.2 m / s

To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is

             Push = v_{e} dM / dt

let's calculate

             Push = 2500  75

             Push = 187500 N

            If we use Newton's second law

             F = m a

             a = F / m

let's calculate

              a = 187500/2250

              a = 83.33 m / s²

7 0
3 years ago
A 0.5 m3 container is filled with a fluid whose specific volume is 0.001 m3/kg. At standard gravitational acceleration, the cont
xenn [34]

Answer: the contents of this container weighs 4905 kg.m/s²

Explanation:

Given that;

volume of a container V = 0.5 m³

we know that standard gravitational acceleration g = 9.81 m/s²

specific volume of liquid filled in the container v = 0.001 m³/kg

now we express the equation for weight of the container.

W = mg

W = (pV)g

W = Vg / ν

so we substitute

W =  (0.5 m³)(9.81 m/s ) / 0.001 m³/kg

W = 4.905 / 0.001

W = 4905 kg.m/s²

Therefore, the contents of this container weighs 4905 kg.m/s²

5 0
3 years ago
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