Answer:
a. give limitations for using a particular lane of traffic
Answer:
Here we have some of the requirement of practical fuel are
1. It must contain large amount of stored energy. So that more amount of power output available to run the engines, motors etc.
2. It must occur in abundance in nature or be easy to produce.
3. The fuel must be made up of elements that combine easily with oxygen. Foe example if hydrogen molecules reacts with oxygen. Then the products are at the reaction of lower energy than the reactants, the result is the explosive release of energy and the product of water.
Answer:
0.6 Ω
Explanation:
As shown in the diagram below,
Since the resistance and the ammeter are connected in series,
(i) The same amount of current flows through them.
(ii) The sum of their individual individual voltage is equal to the total voltage of the circuit.
Applying ohm's law,
V = IR................ Equation 1
Where V = Voltage across the ammeter, I = current flowing through the ammeter, R = resistance of the ammeter.
make R the subject of the equation
R = V/I............... Equation 2
Given: V = 1.2-0.9 = 0.3 V, I = 0.5 A.
Substitute into equation 2
R = 0.3/0.5
R = 0.6 Ω
Set this up as a proportion.
.002 m^3/20 degrees = x/50 degrees
solve for x
x = .005 m^3
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Answer:
21.21 m/s
Explanation:
Let KE₁ represent the initial kinetic energy.
Let v₁ represent the initial velocity.
Let KE₂ represent the final kinetic energy.
Let v₂ represent the final velocity.
Next, the data obtained from the question:
Initial velocity (v₁) = 15 m/s
Initial kinetic Energy (KE₁) = E
Final final energy (KE₂) = double the initial kinetic energy = 2E
Final velocity (v₂) =?
Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:
KE = ½mv²
NOTE: Mass (m) = constant (since we are considering the same car)
KE₁/v₁² = KE₂/v₂²
E /15² = 2E/v₂²
E/225 = 2E/v₂²
Cross multiply
E × v₂² = 225 × 2E
E × v₂² = 450E
Divide both side by E
v₂² = 450E /E
v₂² = 450
Take the square root of both side.
v₂ = √450
v₂ = 21.21 m/s
Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.
