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lord [1]
3 years ago
10

10. Ron’s father finally installed cruise control on their flying car. He was having trouble because he kept of forgetting that

he needed to balance out the opposite force of air resistance on the car (FRICTION) with what below?
A) The force of the driver’s foot
B) The forward force applied by the engine
C) The brakes of the car
Physics
1 answer:
serg [7]3 years ago
7 0

Air drag is the frictional force which is due to air molecules and it is always opposite to the direction of motion

So here direction of air resistance is always opposite to the motion or velocity

Now if Car is travelling then the force on the car due to engine is always in the direction of velocity of car

Because here engine will provide the force to accelerate the car.

So opposite to that motion there is air resistance.

So here correct answer will be

<em>B) The forward force applied by the engine</em>

You might be interested in
What is a charge? how objects can be charged? short answer pls​
nevsk [136]

Answer:

An electrical charge is created when electrons are transferred to or removed from an object. Because electrons have a negative charge, when they are added to an object, it becomes negatively charged. When electrons are removed from an object, it becomes positively charged.

Explanation:

plz Mark my answer in brainlist and follow me

8 0
3 years ago
A 5.00-V battery charges the parallel plates in a capacitor, with a plate area of 865 mm2 and an air-filled separation of 3.00 m
Westkost [7]

Answer:

W = 3.21x10⁻¹¹ J

Explanation:

The work required to separate the plates can be calculated using the following equation:

W = U_{2} - U_{1} = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2})

<u>Where</u>:

U₂: is the final stored energy

U₁: is the initial stored energy

C₂: is the final capacitance

C₁: is the initial capacitance

V₁: is the initial potential difference = 5.00 V

V₂: is the final potential difference

The initial and final capacitance is:

C_{1} = \epsilon_{0}*\frac{A}{d_{1}}

<u>Where</u>:

ε₀: is the vacuum permittivity = 8.85x10⁻¹² C²/(N*m²)

d: is the initial distance = 3.00 mm = 3.00x10⁻³ m    

A: is the plate area = 865 mm² =  8.65x10⁻⁴ m²

C_{1} = \epsilon_{0}*\frac{A}{d_{1}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 \cdot 10^{-3} m} = 2.55 \cdot 10^{-12} F      

Similarly, C₂ is:

C_{2} = \epsilon_{0}*\frac{A}{d_{2}} = 8.85 \cdot 10^{-12} C^{2}/(N*m^{2})*\frac{8.65 \cdot 10^{-4} m^{2}}{3.00 + 3.00 \cdot 10^{-3} m} = 1.28 \cdot 10^{-12} F

Now, V₂ can be calculated by finding the initial charge (q₁):

q_{1} = C_{1}V_{1} = 2.55 \cdot 10^{-12} F*5.00 V = 1.28 \cdot 10^{-11} C

Since, q₁ is equal to q₂, V₂ is:

V_{2} = \frac{q_{2}}{C_{2}} = \frac{1.28 \cdot 10^{-11} C}{1.28 \cdot 10^{-12} F} = 10 V

Finally, we can find the work:

W = \frac{1}{2}(C_{2}V_{2}^{2} - C_{1}V_{1}^{2}) = \frac{1}{2}(1.28 \cdot 10^{-12} F*(10 V)^{2} - 2.55 \cdot 10^{-12} F(5.00 V)^{2}) = 3.21 \cdot 10^{-11} J

Therefore, the work required to separate the plates is 3.21x10⁻¹¹ J.

I hope it helps you!

4 0
3 years ago
A cannon ball launched horizontally with a speed of 20m/s and a baseball dropped off a cliff and it accelerates at a rate of 10m
miskamm [114]
If they both start from the same height, then they both hit the ground at the
same time.  It makes no difference if their horizontal speeds aren't equal.
The cannon ball still accelerates downward at the same rate as the baseball.
8 0
3 years ago
The specific heat of aluminum is 0.90 J/gC . How much heat is given off when 25 grams of aluminum is cooled from 55 C to 25 C?
sp2606 [1]

Answer:

2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/gºC? c=0.90J/g. 9 (2 sigs.)

Explanation:

8 0
3 years ago
Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0
3 years ago
Read 2 more answers
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