Before the fuel is burned, it's chock full of chemical energy. When the fuel is burned, the chemical energy is released, and it escapes in the form of heat and light energy.
Answer:
¨chemical energy¨ is released.
1. How fast is the blue car going 1.8 seconds after it starts?
Recall this kinematic equation:
Vf = Vi + aΔt
Vf is the final velocity.
Vi is the initial velocity.
a is the acceleration.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest)
a = 3.7 m/s² (this is the acceleration between t = 0s and t = 4.4s)
Δt = 1.8 s
Substitute the terms in the equation with the given values and solve for Vf:
Vf = 0 + 3.7×1.8
2. How fast is the blue car going 10.0 seconds after it starts?
The car stops accelerating after t = 4.4s and continues at a constant velocity for the next 8.3 seconds. This means the car is traveling at a constant velocity between t = 4.4s and t = 12.7s. At t = 10s the car is still traveling at this constant velocity.
We must use the kinematic equation from the previous question to solve for this velocity. Use the same values except Δt = 4.4s which is the entire time interval during which the car is accelerating:
Vf = 0 + 3.7×4.4
Vf = 16.28 m/s
3. How far does the blue car travel before its brakes are applied to slow down?
We must break down the car's path into two parts: When it is traveling under constant acceleration and when it is traveling at constant velocity.
Traveling under constant acceleration:
Recall this kinematic equation:
d = ×Δt
d is the distance traveled.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 16.28 m/s (determined from question 2).
Δt = 4.4 s
Substitute the terms in the equation with the given values and solve for d:
d = ×4.4
d = 35.8 m
Traveling at constant velocity:
Recall the relationship between velocity and distance:
d = vΔt
d is the distance traveled.
v is the velocity.
Δt is the amount of elapsed time.
Given values:
v = 16.28 m/s (the constant velocity from question 2).
Δt = 8.3 s (the time interval during which the car travels at constant velocity)
Substitute the terms in the equation with the given values:
d = 16.28×8.3
d = 135.1 m
Add up the distances traveled.
d = 35.8 + 135.1
4. What is the acceleration of the blue car once the brakes are applied?
Recall this kinematic equation:
Vf²=Vi²+2ad
Vf is the final velocity.
Vi is the initial velocity.
a is the acceleration
d is the distance traveled.
Given values:
Vi = 16.28 m/s
Vf = 0 m/s
d = 216 m - 170.9 m = 45.1 m (subtracting the distance already traveled from the total path length)
Substitute the terms in the equation with the given values and solve for a:
0² = 16.28²+2a×45.1
5. What is the total time the blue car is moving?
We already know the time during which the car is traveling under constant acceleration and traveling at constant velocity. We now need to solve for the amount of time during which the car is decelerating.
Recall again:
d = ×Δt
Given values:
d = 45.1 m
Vi = 16.28 m/s (the velocity the car was traveling at before hitting the brakes).
Vf = 0 m/s (the car slows to a stop).
Substitute the terms in the equation with the given values and solve for Δt:
45.1 = ×Δt
Δt = 5.54s
Add up the times to get the total travel time:
t = 4.4 + 8.3 + 5.54 =
6. What is the acceleration of the yellow car?
Recall this kinematic equation:
d = ViΔt + 0.5aΔt²
d is the distance traveled.
Vi is the initial velocity.
a is the acceleration.
Δt is the amount of elapsed time.
Given values:
d = 216 m (both cars meet at 216m)
Vi = 0 m/s (the car starts at rest)
Δt = 18.24 s (take the same amount of time to reach 216m)
Substitute the terms in the equation with the given values and solve for a:
216 = 0×18.24 + 0.5a×18.24²
b. in moderate amounts.
c. without a specific timeline.
d. shouldn’t match your personal tastes.
B.the speed of the car always increases, giving it greater force since the mass will remain the same.
C.the car now has a greater overall mass - its own mass plus the mass of the object by which it has been struck.
D.the people in the car will continue to move in the same direction and at the same speed as before the impact.
When an impact causes the car to suddenly change its motion, (D) the people in the car will continue to move in the same direction and at the same speed as before the impact.
Newton's first law of motion states that Unless influenced by an imbalanced force, a body at rest stays at rest, and a body in motion keeps moving in a straight path at a constant pace. This law is commonly known as law of inertia and the resistance of a body to maintain its state of rest is called inertia of rest, same as, the resistance of a body to maintain its state of motion is called inertia of motion.
When a car is moving with certain speed, the car along with the people inside the car is in inertia of motion. In an event of accident, the car comes in rest in no time but the man in the car is still in inertia of motion. So, his body moves forward in the car which may cause injuries. To prevent such injuries, the people inside a car are advised to use safety belts.
So, correct answer is option (D).
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Answer:
D :)
Explanation:
Explain why ultra-high voltages are
used to carry electricity over transmission lines.
A. Commutator
B. Loop of wire
C. Permanent magnets
D. Battery
PLS HELP
Commutator is part of a motor causes its electromagnet to turn in the same direction consistently, transforming electrical energy into mechanical energy.
Energy is the ability or capability to do tasks, such as the ability to move an item (of a certain mass) by exerting force. Energy can exist in many different forms, including electrical, mechanical, chemical, thermal, or nuclear, and it can change its form.
A commutator is a rotary electrical switch in certain types of electric motors and electrical generators.
Commutator is part of a motor causes its electromagnet to turn in the same direction consistently, transforming electrical energy into mechanical energy.
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Answer:
A. Commutator
Explanation: