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alina1380 [7]
3 years ago
8

Consider the relationship of the variables in Newton’s second law. In a drag car race, the force applied to the car is doubled b

y the driver stepping on the gas pedal. The acceleration of the car will . The mass of the car will . The velocity of the car will .
Physics
2 answers:
kkurt [141]3 years ago
5 0

When the driver mashes the gas, puts the pettle to the mettle, and doubles the engine force applied to the car . . .

-- the acceleration of the car doubles

-- the mass of the car doesn't change

-- the velocity of the car starts to increase in the direction in which the car is already moving.  If the car's speed was already increasing forward before the force adjustment, then the rate at which it increases due to the doubled force will double.

daser333 [38]3 years ago
4 0

Recall that the force on an object is related to the mass and acceleration of that object by the formula F = ma, where m is the mass of the object and a is its acceleration. What happens when we double F? Well, you might remember from algebra that, in order to keep our equality true, if we double one side, we must also double the other, so our equation becomes 2F = 2ma. Now, this means one of two things: either the mass has doubled, or the acceleration has doubled.

We can tell right away that it'd be absurd if a race car doubled in mass every time it hit the gas, so the quantity doubling must be the <em>acceleration. </em>If we call the car's current velocity v1, we'll be adding the doubled acceleration to get its new velocity. Mathematically, v = v1 + 2a.

We can now conclude that, by doubling the force:

  • The acceleration of the car will double,
  • The mass of the car will stay the same, and
  • The velocity of the car will increase by double the original acceleration
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Work and Power Practice Calculations
Elza [17]

Answer:

1. W = F d = 20 N * 6 m = 120 J

2. F = W / d = 60 J / 2 m = 30 N

3. d = W / F = 350 J / 85 N = 4.12 m

4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts

5. W = P t = 500 W * 120 sec = 60,000 J

6. t = W / P = 550 J / 310 W = 1.77 sec

5 0
2 years ago
Pete is driving down 7th Street. He drives 300 meters in 18 seconds. Assuming he does not speed up or slow down, what is his spe
Korolek [52]

Answer:

16.67m/s

Explanation:

Given parameters:

Distance Pete drove  = 300m

Time taken  = 18s

Unknown:

Speed  = ?

Solution:

Speed is the distance traveled per unit of time.

It is mathematically expressed as;

   Speed  = \frac{distance}{time}

Insert the parameters and solve;

  Speed  = \frac{300}{18}  = 16.67m/s

3 0
3 years ago
There is little vertical air movement in the __ because of the cold, heavy air at the bottom of the layer.
pogonyaev
The correct answer is the Mesosphere. <span>Mesosphere </span><span>is the layer of the Earth's atmosphere that is directly above the stratopause and directly below the mesopause.</span>
4 0
3 years ago
Read 2 more answers
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
What is the difference between compound, element and mixture?
defon
Compound; consists of atoms of two or more different elements bound together,can be broken down into a simpler type of matter (elements) by chemical means (but not by physical means) has properties that are different from its component elements, and always contains the same ratio of its component atoms.Mixtures; Note that a mixture:consists of two or more different elements and/or compounds physically intermingled, can be separated into its components by physical means, and often retains many of the properties of its components.
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3 years ago
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